#include <cstdio>

template <typename T> inline bool chkmax(T &x, T y) {
  return x<y? x=y, true : false;
}

enum {
  MAXN = 3000 + 5,
};

const int MOD = 1000000007;

int n, m;
char s[MAXN];

int dp[MAXN][MAXN];
int pre[MAXN];
int R[MAXN];

void clear() {
  int i;
  for (i=0; i<MAXN; ++i)
    R[i] = i+1;
}

void exec() {
  int i, l, r, nj, j;
  clear();
  scanf("%d%d%s", &n, &m, s);
  for (i=0; i<m; ++i) {
    scanf("%d%d", &l, &r), --l;
    chkmax(R[l], r);
  }

  for (i=1; i<n; ++i)
    chkmax(R[i], R[i-1]);

  R[n] = n;
  for (i=0; i<n; ++i)
    pre[i+1] = pre[i] + (s[i] == '1');

  for (i=0; i<=R[0]; ++i)
    dp[1][i] = (pre[R[0]] == i);
  for (i=1; i<=n; ++i) {
    for (j=1; j<=R[i-1]-(i-1); ++j) {
      nj = j + pre[R[i]] - pre[R[i-1]] - 1;
      (dp[i+1][nj] += dp[i][j]) %= MOD;
    }
    for (j=0; j< R[i-1]-(i-1); ++j) {
      nj = j + pre[R[i]] - pre[R[i-1]];
      (dp[i+1][nj] += dp[i][j]) %= MOD;
    }
  }
  printf("%d\n", dp[n+1][0]);
}

int main() {
  if (fopen("F.in", "r") != NULL) {
    freopen("F.in", "r", stdin);
    freopen("F.out", "w", stdout);
  }
  exec();
  return 0;
}

### Prompt

Please formulate a cpp solution to the following problem:

There is a string S of length N consisting of characters `0` and `1`. You will perform the following operation for each i = 1, 2, ..., m:

* Arbitrarily permute the characters within the substring of S starting at the l_i-th character from the left and extending through the r_i-th character.



Here, the sequence l_i is non-decreasing.

How many values are possible for S after the M operations, modulo 1000000007(= 10^9+7)?

Constraints

* 2≦N≦3000
* 1≦M≦3000
* S consists of characters `0` and `1`.
* The length of S equals N.
* 1≦l_i < r_i≦N
* l_i ≦ l_{i+1}

Input

The input is given from Standard Input in the following format:


N M
S
l_1 r_1
:
l_M r_M


Output

Print the number of the possible values for S after the M operations, modulo 1000000007.

Examples

Input

5 2
01001
2 4
3 5


Output

6


Input

9 3
110111110
1 4
4 6
6 9


Output

26


Input

11 6
00101000110
2 4
2 3
4 7
5 6
6 10
10 11


Output

143

### Response

```cpp
#include <cstdio>

template <typename T> inline bool chkmax(T &x, T y) {
  return x<y? x=y, true : false;
}

enum {
  MAXN = 3000 + 5,
};

const int MOD = 1000000007;

int n, m;
char s[MAXN];

int dp[MAXN][MAXN];
int pre[MAXN];
int R[MAXN];

void clear() {
  int i;
  for (i=0; i<MAXN; ++i)
    R[i] = i+1;
}

void exec() {
  int i, l, r, nj, j;
  clear();
  scanf("%d%d%s", &n, &m, s);
  for (i=0; i<m; ++i) {
    scanf("%d%d", &l, &r), --l;
    chkmax(R[l], r);
  }

  for (i=1; i<n; ++i)
    chkmax(R[i], R[i-1]);

  R[n] = n;
  for (i=0; i<n; ++i)
    pre[i+1] = pre[i] + (s[i] == '1');

  for (i=0; i<=R[0]; ++i)
    dp[1][i] = (pre[R[0]] == i);
  for (i=1; i<=n; ++i) {
    for (j=1; j<=R[i-1]-(i-1); ++j) {
      nj = j + pre[R[i]] - pre[R[i-1]] - 1;
      (dp[i+1][nj] += dp[i][j]) %= MOD;
    }
    for (j=0; j< R[i-1]-(i-1); ++j) {
      nj = j + pre[R[i]] - pre[R[i-1]];
      (dp[i+1][nj] += dp[i][j]) %= MOD;
    }
  }
  printf("%d\n", dp[n+1][0]);
}

int main() {
  if (fopen("F.in", "r") != NULL) {
    freopen("F.in", "r", stdin);
    freopen("F.out", "w", stdout);
  }
  exec();
  return 0;
}

```