n个人在w*h的监狱里面想要逃跑,已知他们的同伙在坐标(bi,h)接应他们,他们现在被关在(ai,1)现在他们必须要到同伙那里才有逃出去的机会,这n个人又很蠢只会从(x,y)->(x+1,y),(x,y+1)并且这他们走过的路径不能相交如果相交第一个经过后就会有第二个人经过时候就会有一名狱警在那等他,第二个人就会被抓,假设他们不会同时踩到某个格子，那么他们的逃跑路线有多少不同的方案数。如果两个方案不同那么存在一个人踩的格子至少有一个是另外一个方案的没踩过

第一行一个t(t<=20)表示测试样例 
第二行两个3个正整数n,w,h(n<=100,w,h<=1e9) 
接下来n行每行两个整数 
ai,bi(ai,bi<=w) 

输出一个整数表示答案最终结果取膜109*1000003

1
2 4 2
1 2
3 4

4

ll w = M / m[i];
d = exgcd(m[i], w, x, y);
ret = (ret + modmul(modmul(y, w, M), a[i], M) ) % M;

#include<set>
#include<map>
#include<stack>
#include<cmath>
#include<queue>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
typedef long long ll;
using namespace std;
const int maxn = 1e5 + 10;
const int seed = 131;
const ll MOD = 109 * 1000003;
const int INF = 0x3f3f3f3f;
int a[105], b[105];
ll e[105][105];
ll fac[2][1000010], inv[2][1000010];
ll modmul(ll a, ll b, ll p){ll ret = 0;while(b) {if(b & 1) ret = ret + a;if(ret >= p) ret -= p;a <<= 1;if(a >= p) a -= p;b >>= 1;}return ret;
}
ll pmul(ll a, ll b, ll p){ll ans = 1;a %= p;while(b){if(b & 1) ans = ans * a % p;a = a * a % p;b >>= 1;}return ans;
}
ll C(ll n, ll m, ll p, int i){if(m > n) return 0;return fac[i][n] * inv[i][m] * inv[i][n - m] % p;
}
ll Lucas(ll n, ll m, ll p, int i){if(m == 0) return 1;if(n < p && m < p) return C(n, m, p, i);return C(n % p, m % p, p, i) * Lucas(n / p, m / p, p, i) % p;
}
ll mm[2] = {56, 486240};
ll remainder(ll a[], ll m[], int len){ll x, y, ret = 0;ll M = MOD;for (int i = 0; i < len; i++){ll w = M / m[i];ret = (ret + a[i] * mm[i] * w) % M;}return ret;
}
ll guass(int n, ll p){ll ans = 1, f = 1;for(int i = 1; i <= n; i++){for(int j = i + 1; j <= n; j++){int x = i, y = j;while(e[y][i]){ll t = e[x][i] / e[y][i];for(int k = i; k <= n; k++)e[x][k] = (e[x][k] - e[y][k] * t % p) % p;swap(x,y);}if(x != i){for(int k = 1; k <= n; k++)swap(e[i][k], e[j][k]);f = -f;}}ans = ans * e[i][i] % p;if(ans == 0) return 0;}return (ans * f + p) % p;
}
void init(int x, int n){fac[x][0] = 1;for (ll i = 1; i < n; i++) fac[x][i] = fac[x][i - 1] * i % n;inv[x][n - 1] = pmul(fac[x][n - 1], n - 2, n);for (ll i = n - 2; i >= 0; i--) inv[x][i] = inv[x][i + 1] * (i + 1) % n;
}
ll lucas[2];
ll pp[2] = {109, 1000003};
ll solve(ll n, ll m){ll ret;for(int i = 0; i < 2; i++){lucas[i] = Lucas(n, m, pp[i], i);}ret = remainder(lucas, pp, 2);return ret;
}
int main(){init(0, 109);init(1, 1000003);int t;scanf("%d", &t);while(t--){ll n, w, h;scanf("%lld%lld%lld", &n, &w, &h);for(int i = 1; i <= n; i++)scanf("%d%d", &a[i], &b[i]);for(int i = 1; i <= n; i++){for(int j = 1; j <= n; j++){if(b[j] >= a[i]){e[i][j] = solve(h - 1 + b[j] - a[i], b[j] - a[i]);}else e[i][j] = 0;}}printf("%lld\n", guass(n, MOD));}return 0;
}

#include<set>
#include<map>
#include<stack>
#include<cmath>
#include<queue>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
typedef long long ll;
using namespace std;
const int maxn = 1e5 + 10;
const int seed = 131;
const ll MOD = 109 * 1000003;
const int INF = 0x3f3f3f3f;
ll a[maxn], b[maxn];
ll e[105][105];
ll prime[maxn], p[maxn], pn;
ll fac[2][1000010];
ll pmul(ll a, ll b, ll p){ll ans = 1;while(b){if(b & 1) ans = ans * a % p;a = a * a % p;b >>= 1;}return ans;
}
ll modmul(ll a, ll b, ll p) {ll ret = 0;while(b) {if(b & 1) ret = (ret + a) % p;a = (a + a) % p;b >>= 1;}return ret;
}
ll Lucas(ll n, ll m, ll p, int i) {ll ret=1;while(n && m) {ll a = n%p, b = m%p;if(a<b) return 0;ret = (ret * fac[i][a] * pmul(fac[i][b]*fac[i][a - b] % p, p-2, p)) % p;n/=p;m/=p;}return ret;
}ll exgcd (ll a, ll b, ll &x, ll &y) {if (!b) {x = 1, y = 0;return a;}int ans = exgcd ( b , a % b , y , x );y -= a / b * x;return ans;
}
ll remainder(ll a[], ll m[], int len) {ll d, x, y, ret = 0;ll M = 1;for (int i = 0; i < len; i++) M *= m[i];for (int i = 0; i < len; i++) {ll w = M / m[i];d = exgcd(m[i], w, x, y);ret = (ret + modmul(modmul(y, w, M), a[i], M) ) % M;}return (ret + M) % M;
}
ll guass(int n, ll MOD){ll ans = 1, f = 1;for(int i = 1; i <= n; i++){for(int j = i + 1; j <= n; j++){int x = i, y = j;while(e[y][i]){ll t = e[x][i] / e[y][i];for(int k = i; k <= n; k++)e[x][k] = (e[x][k] - e[y][k] * 1LL * t % MOD) % MOD;swap(x,y);}if(x != i){for(int k = 1; k <= n; k++)swap(e[i][k], e[j][k]);f = -f;}}ans = ans * e[i][i] % MOD;if(ans == 0) return 0;}return (ans * f + MOD) % MOD;
}
void init(){memset(prime, 0, sizeof(prime));pn = 0;for(ll i = 2; i < maxn; i++){if(!prime[i]){p[pn++] = i;for(ll j = i * i; j < maxn; j += i)prime[i] = 1;}}fac[0][0] = 1;for(int i = 1; i <= 109; i++){fac[0][i] = (fac[0][i-1]*i) % 109;}fac[1][0] = 1;for(ll i = 1; i <= 1000003; i++){fac[1][i] = (fac[1][i-1]*i) % 1000003;}
}
ll solve(ll n, ll m){ll ret;ll lucas[2];ll p[2] = {109, 1000003};for(int i = 0; i < 2; i++){lucas[i] = Lucas(n, m, p[i], i);}return ret = remainder(lucas, p, 2);
}
int main(){init();int t;scanf("%d", &t);while(t--){ll n, w, h;scanf("%lld%lld%lld", &n, &w, &h);for(int i = 1; i <= n; i++)scanf("%lld%lld", &a[i], &b[i]);for(int i = 1; i <= n; i++){for(int j = 1; j <= n; j++){if(b[j] >= a[i]){e[i][j] = solve(h - 1 + b[j] - a[i], b[j] - a[i]);}else e[i][j] = 0;}}printf("%lld\n", guass(n, MOD));}return 0;
}