#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <string>
#include <iostream>
#include <queue>
#include <map>
#include <cmath>typedef long long ll;
const int N = 1e5 + 5;
const int INF = 0x3f3f3f3f;int main(void)  {int T; std::cin >> T;while (T--) {ll a, b, c;    std::cin >> a >> b >> c;ll d = b * b - 4 * 1ll * a * c;if (d < 0) puts ("NO");else  {ll e = sqrt (d);if (e * e == d) puts ("YES");else puts ("NO");}}return 0;
}
/*
long long   long double
sqrt () HDU %I64d
*/

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <string>
#include <queue>
#include <map>
#include <iostream>
#include <cmath>int dp[22][2002];int main(void)   {int T; scanf ("%d", &T);while (T--)  {int n; scanf ("%d", &n);memset (dp, 0, sizeof (dp));dp[0][0] = 1;for (int w, i=1; i<=n; ++i) {scanf ("%d", &w);for (int j=0; j<=2000; ++j)  {if (!dp[i-1][j])   continue;dp[i][j] = dp[i][j+w] = 1;dp[i][abs (j-w)] = 1;}}int m;    scanf ("%d", &m);for (int w, i=1; i<=m; ++i)   {scanf ("%d", &w);if (dp[n][w])   puts ("YES");else puts ("NO");}}return 0;
}
//01dp "twice"

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <string>
#include <queue>
#include <map>
#include <cmath>typedef long long ll;
const int N = 5e2 + 5;
const int MOD = 5201314;
int dp[2][N][N];
char str[N][N];void add(int &a, int b)  {a += b;if (a >= MOD) a %= MOD;
}int main(void) {int T; scanf ("%d", &T);while (T--)  {int n; scanf ("%d", &n);for (int i=1; i<=n; ++i)  scanf ("%s", str[i] + 1);if (str[1][1] != str[n][n])    puts ("0");else   {memset (dp, 0, sizeof (dp));int now = 0;dp[now][1][n] = 1;for (int s=1; s<n; ++s)  {now ^= 1; memset (dp[now], 0, sizeof (dp[now]));for (int x1=1; x1<=n; ++x1)    {for (int x2=n; x2>=1; --x2)   {if (x1 - 1 > s || n - x2 > s)    continue;int y1 = 1 + s - (x1 - 1);int y2 = n - (s - (n - x2));if (str[x1][y1] != str[x2][y2])  continue;add (dp[now][x1][x2], dp[now^1][x1][x2]);add (dp[now][x1][x2], dp[now^1][x1][x2+1]);add (dp[now][x1][x2], dp[now^1][x1-1][x2]);add (dp[now][x1][x2], dp[now^1][x1-1][x2+1]);}}}int ans = 0;for (int i=1; i<=n; ++i)  add (ans, dp[now][i][i]);printf ("%d\n", ans);}}return 0;
}