/*T1 求多项式结果的后k位，由于k<=8， 所以不必考虑高精了，直接把多项式一遍算，一边取模，注意用快速幂，最后补0输出即可。复杂度 O (n log b)
*/
#include <cstdio>#define Max 100090void read (int &now)
{register char word = getchar ();bool temp = false;for (now = 0; word < '0' || word > '9'; word = getchar ())if (word == '-')temp = true;for (; word >= '0' && word <= '9'; now = now * 10 + word - '0', word = getchar ());if (temp)now = -now;
}int Mod = 1;
int N, K, X;int a[Max], b[Max];long long Fast_Pow (long long x, long long p)
{register long long res = 1;for (; p; p >>= 1){if (p & 1)res = res * x % Mod;x = x % Mod * x % Mod;}return res;
}#define Judge
long long Answer;int number[Max / 1000];int main (int argc, char *argv[])
{#ifdef Judgefreopen ("digits.in", "r", stdin);freopen ("digits.out", "w", stdout);#endif read (N);read (K);for (int i = 1; i <= N; i ++){read (a[i]);read (b[i]);}for (int i = K, res = 10; i; i >>= 1){if (i & 1)Mod = Mod * res;res *= res;}read (X);for (int i = 1; i <= N; i ++)Answer = (Answer + (a[i] * Fast_Pow (X, b[i]) % Mod)) % Mod; int Count = 0;for (; Answer > 0; number[++ Count] = Answer % 10, Answer /= 10);for (int i = Count; i < K; i ++, printf ("0\n"));for (; Count >= 1; printf ("%d\n", number[Count --]));return 0;
} 

/*T2把多项式左边的负项移到右边， O(N^3)预处理出右边的结果，用一个桶记录下来，后O(N^3)枚举左边的结果， 加减就好了 复杂度O(N^3)
*/
#include <cstdio>#define Max 60000002#define N 6void read (int &now)
{register char word = getchar ();bool temp = false;for (now = 0; word < '0' || word > '9'; word = getchar ())if (word == '-')temp = true;for (; word >= '0' && word <= '9'; now = now * 10 + word - '0', word = getchar ());if (temp)now = -now;
}#define Judgeint K;short z_count[Max];
short f_count[Max];int a1, a2, a3, a4, a5, a6;int main (int argc, char *argv[])
{#ifdef Judge freopen ("equation.in", "r", stdin);freopen ("equation.out", "w", stdout);#endifread (K);read (a1);read (a2);read (a3);read (a4);read (a5);read (a6);    register int x;for (register int i = 1, j, k; i <= K; i ++)for (j = 1; j <= K; j ++)for (k = 1; k <= K; k ++){x = a2 * i + a4 * j + a6 * k;if (x >= 0)z_count[x] ++;elsef_count[-x] ++;}int Answer = 0;for (register int i = 1, j, k; i <= K; i ++)for (j = 1; j <= K; j ++)for (k = 1; k <= K; k ++){x = a1 * i + a3 * j + a5 * k;if (x >= 0)Answer += z_count[x];elseAnswer += f_count[-x];}printf ("%d", Answer);return 0;
}

/*T3 由于k>=n，所以一天最多走一条边，则问题转化为了求最小瓶颈生成树的裸题。。。。若一直到最后起点与终点都不联通，那么则无解复杂度 O(M logM + M) = O (M log M)
*/
#include <algorithm>
#include <cstdio>#define Max 8000void read (int &now)
{register char word = getchar ();for (now = 0; word < '0' || word > '9'; word = getchar ());for (; word >= '0' && word <= '9'; now = now * 10 + word - '0', word = getchar ());
}inline int max (int a, int b)
{return a > b ? a : b;
}struct Edge
{int from;int to;int dis;bool operator < (const Edge &now) const{return this->dis < now.dis;}
};class Unio_Find_Set
{private :int father[Max];public :void Prepare (int N){for (int i = 1; i <= N; i ++)father[i] = i;}int Find (int x){return father[x] == x ? x : father[x] = Find (father[x]);}inline void Unio (int a, int b){father[a] = b;}
};Unio_Find_Set Ufs;int N, M, K;Edge edge[Max * 30];
#define Judgeint main (int argc, char *argv[])
{#ifdef Judge freopen ("graph.in", "r", stdin);freopen ("graph.out", "w", stdout);#endifread (N);read (M);read (K);for (int i = 1; i <= M; i ++){read (edge[i].from);read (edge[i].to);read (edge[i].dis);}Ufs.Prepare (N);std :: sort (edge + 1, edge + 1 + M);int Count = 0;for (register int i = 1, x, y; i <= M; i ++){x = Ufs.Find (edge[i].from);y = Ufs.Find (edge[i].to);if (x != y){Ufs.Unio (x, y); Count ++;}if (Ufs.Find (1) == Ufs.Find (N)){printf ("%d", edge[i].dis);return 0;}if (Count == N - 1)break;}printf ("-1");return 0;
}