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;
template <typename T, typename U>
inline void smin(T &a, U b) {
  if (a > b) a = b;
}
template <typename T, typename U>
inline void smax(T &a, U b) {
  if (a < b) a = b;
}
template <typename T>
inline void gn(T &first) {
  char c, sg = 0;
  while (c = getchar(), (c > '9' || c < '0') && c != '-')
    ;
  for ((c == '-' ? sg = 1, c = getchar() : 0), first = 0; c >= '0' && c <= '9';
       c = getchar())
    first = (first << 1) + (first << 3) + c - '0';
  if (sg) first = -first;
}
template <class T, class T1>
inline void gn(T &first, T1 &second) {
  gn(first);
  gn(second);
}
template <class T, class T1, class T2>
inline void gn(T &first, T1 &second, T2 &z) {
  gn(first);
  gn(second);
  gn(z);
}
template <typename T>
inline void print(T first) {
  if (first < 0) {
    putchar('-');
    return print(-first);
  }
  if (first < 10) {
    putchar('0' + first);
    return;
  }
  print(first / 10);
  putchar(first % 10 + '0');
}
template <typename T>
inline void println(T first) {
  print(first), putchar('\n');
}
template <typename T>
inline void printsp(T first) {
  print(first), putchar(' ');
}
template <class T, class T1>
inline void print(T first, T1 second) {
  printsp(first), println(second);
}
template <class T, class T1, class T2>
inline void print(T first, T1 second, T2 z) {
  printsp(first), printsp(second), println(z);
}
int power(int a, int b, int m, int ans = 1) {
  for (; b; b >>= 1, a = 1LL * a * a % m)
    if (b & 1) ans = 1LL * ans * a % m;
  return ans;
}
int dp[155][155][155][2][2];
int sL[155][155][155][2], sR[155][155][155][2], ss[155][155][155],
    so[155][155][155][2][2];
int ssL[155][155][2], ssR[155][155][2];
inline void add(int &a, int b) {
  a += b;
  if (a >= 1000000007) a -= 1000000007;
}
int main() {
  int n, m;
  gn(n, m);
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    for (int L = 1; L <= m; L++) {
      for (int R = L; R <= m; R++) {
        dp[i][L][R][0][0] = (ss[i - 1][L][R] + 1) % 1000000007;
        dp[i][L][R][0][1] =
            ((long long)sL[i - 1][L][R][1] + sL[i - 1][L][R][0] -
             ss[i - 1][L][R] + ss[i - 1][L][R - 1] + 1000000007) %
            1000000007;
        dp[i][L][R][1][0] =
            ((long long)sR[i - 1][L][R][1] + sR[i - 1][L][R][0] -
             ss[i - 1][L][R] + ss[i - 1][L + 1][R] + 1000000007) %
            1000000007;
        dp[i][L][R][1][1] =
            ((long long)so[i - 1][L - 1][R + 1][0][0] +
             so[i - 1][L - 1][R][0][1] + so[i - 1][L][R + 1][1][0] +
             so[i - 1][L][R][1][1]) %
            1000000007;
        add(ans, dp[i][L][R][0][0]);
        add(ans, dp[i][L][R][0][1]);
        add(ans, dp[i][L][R][1][0]);
        add(ans, dp[i][L][R][1][1]);
      }
    }
    for (int L = 1; L <= m; L++) {
      for (int R = m; R >= L; R--)
        for (int a = 0; a < 2; a++)
          ssL[L][R][a] = (ssL[L][R + 1][a] + dp[i][L][R][0][a]) % 1000000007;
    }
    for (int R = 1; R <= m; R++) {
      for (int L = 1; L <= R; L++)
        for (int a = 0; a < 2; a++)
          ssR[L][R][a] = (ssR[L - 1][R][a] + dp[i][L][R][a][0]) % 1000000007;
    }
    for (int len = 1; len <= m; len++) {
      for (int L = 1, R = L + len - 1; R <= m; L++, R++) {
        ss[i][L][R] = ((long long)dp[i][L][R][0][0] + ss[i][L + 1][R] +
                       ss[i][L][R - 1] - ss[i][L + 1][R - 1] + 1000000007) %
                      1000000007;
        for (int a = 0; a < 2; a++) {
          sL[i][L][R][a] = (sL[i][L + 1][R][a] + ssL[L][R][a]) % 1000000007;
          sR[i][L][R][a] = (sR[i][L][R - 1][a] + ssR[L][R][a]) % 1000000007;
        }
      }
    }
    for (int len = m; len >= 1; len--) {
      for (int L = 1, R = L + len - 1; R <= m; L++, R++) {
        for (int a = 0; a < 2; a++)
          for (int b = 0; b < 2; b++) {
            so[i][L][R][a][b] = ((long long)dp[i][L][R][a][b] +
                                 so[i][L - 1][R][a][b] + so[i][L][R + 1][a][b] -
                                 so[i][L - 1][R + 1][a][b] + 1000000007) %
                                1000000007;
          }
      }
    }
  }
  println(ans);
  return 0;
}
