Dima loves making pictures on a piece of squared paper. And yet more than that Dima loves the pictures that depict one of his favorite figures. 

A piece of squared paper of size n × m is represented by a table, consisting of n rows and m columns. All squares are white on blank squared paper. Dima defines a picture as an image on a blank piece of paper, obtained by painting some squares black.

The picture portrays one of Dima's favorite figures, if the following conditions hold:

  * The picture contains at least one painted cell; 
  * All painted cells form a connected set, that is, you can get from any painted cell to any other one (you can move from one cell to a side-adjacent one); 
  * The minimum number of moves needed to go from the painted cell at coordinates (x1, y1) to the painted cell at coordinates (x2, y2), moving only through the colored cells, equals |x1 - x2| + |y1 - y2|. 



Now Dima is wondering: how many paintings are on an n × m piece of paper, that depict one of his favorite figures? Count this number modulo 1000000007 (109 + 7).

Input

The first line contains two integers n and m — the sizes of the piece of paper (1 ≤ n, m ≤ 150).

Output

In a single line print the remainder after dividing the answer to the problem by number 1000000007 (109 + 7).

Examples

Input

2 2


Output

13


Input

3 4


Output

571
#include <bits/stdc++.h>
using namespace std;
int n, m;
long long ret, cnt, dp[155][155][155][2][2], S[2][2][155][155];
long long MOD = 1000000007;
inline void Mod(long long &a) {
  while (a >= MOD) a -= MOD;
}
inline long long Sum(int a, int b, int l1, int h1, int l2, int h2) {
  if (l2 > h2 || l1 > h1) return 0;
  long long ret = 0;
  ret += S[a][b][h1 + 1][h2 + 1];
  ret -= S[a][b][l1][h2 + 1];
  ret -= S[a][b][h1 + 1][l2];
  ret += S[a][b][l1][l2];
  return (ret % MOD + MOD) % MOD;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = (0); i < (m); i++)
    for (int j = (i); j < (m); j++) dp[1][i][j][0][0] = 1;
  for (int i = (1); i <= (n); i++) {
    cnt = 0;
    memset(S, 0, sizeof(S));
    for (int j = (0); j < (m); j++)
      for (int k = (j); k < (m); k++)
        for (int a = (0); a < (2); a++)
          for (int b = (0); b < (2); b++) {
            cnt += dp[i][j][k][a][b];
            S[a][b][j + 1][k + 1] = dp[i][j][k][a][b];
          }
    cnt = (cnt % MOD * (n - i + 1)) % MOD;
    ret = (ret + cnt % MOD) % MOD;
    for (int a = (0); a < (2); a++)
      for (int b = (0); b < (2); b++)
        for (int j = (1); j <= (m); j++)
          for (int k = (1); k <= (m); k++) {
            S[a][b][j][k] += S[a][b][j - 1][k] + S[a][b][j][k - 1] -
                             S[a][b][j - 1][k - 1] + MOD;
            Mod(S[a][b][j][k]);
          }
    for (int j = (0); j < (m); j++)
      for (int k = (j); k < (m); k++) {
        dp[i + 1][j][k][0][0] = Sum(0, 0, j, k, j, k);
        dp[i + 1][j][k][1][0] =
            Sum(0, 0, 0, j - 1, j, k) + Sum(1, 0, 0, j, j, k);
        dp[i + 1][j][k][0][1] =
            Sum(0, 0, j, k, k + 1, m - 1) + Sum(0, 1, j, k, k, m - 1);
        dp[i + 1][j][k][1][1] =
            Sum(0, 0, 0, j - 1, k + 1, m - 1) + Sum(1, 0, 0, j, k + 1, m - 1) +
            Sum(0, 1, 0, j - 1, k, m - 1) + Sum(1, 1, 0, j, k, m - 1);
        for (int a = (0); a < (2); a++)
          for (int b = (0); b < (2); b++) Mod(dp[i + 1][j][k][a][b]);
      }
  }
  printf("%I64d\n", ret);
  return 0;
}
