#include <bits/stdc++.h>

using namespace std;

const int maxn = 3005,mo = int(1e9) + 7;

string s;
int a[maxn],m,far[maxn],f[maxn][maxn],co[maxn],n,ans;

int dfs(int l,int r,int c)
{
	if (l == n + 1) return 1;
	if (r < far[l]) c += co[far[l]] - co[r],r = far[l];
	if (~f[l][c]) return f[l][c];
	int ans = 0;
	if (c) ans = (ans + dfs(l + 1,r,c - 1)) % mo;
	if (r - l + 1 - c) ans = (ans + dfs(l + 1,r,c)) % mo;
	return f[l][c] = ans;
}

int main()
{
	cin >> n >> m;
	cin >> s;
	for(int i = 1;i <= n;i ++) a[i] = s[i - 1] - '0',co[i] = co[i - 1] + a[i];
	for(int i = 1,l,r;i <= m;i ++) cin >> l >> r,far[l] = max(far[l],r);
	for(int i = 1;i <= n;i ++) far[i] = max(far[i],far[i - 1]),far[i] = max(far[i],i);
	memset(f,255,sizeof f);
	cout << dfs(1,0,0) << endl;
	return 0;
}

### Prompt

Your task is to create 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 <bits/stdc++.h>

using namespace std;

const int maxn = 3005,mo = int(1e9) + 7;

string s;
int a[maxn],m,far[maxn],f[maxn][maxn],co[maxn],n,ans;

int dfs(int l,int r,int c)
{
	if (l == n + 1) return 1;
	if (r < far[l]) c += co[far[l]] - co[r],r = far[l];
	if (~f[l][c]) return f[l][c];
	int ans = 0;
	if (c) ans = (ans + dfs(l + 1,r,c - 1)) % mo;
	if (r - l + 1 - c) ans = (ans + dfs(l + 1,r,c)) % mo;
	return f[l][c] = ans;
}

int main()
{
	cin >> n >> m;
	cin >> s;
	for(int i = 1;i <= n;i ++) a[i] = s[i - 1] - '0',co[i] = co[i - 1] + a[i];
	for(int i = 1,l,r;i <= m;i ++) cin >> l >> r,far[l] = max(far[l],r);
	for(int i = 1;i <= n;i ++) far[i] = max(far[i],far[i - 1]),far[i] = max(far[i],i);
	memset(f,255,sizeof f);
	cout << dfs(1,0,0) << endl;
	return 0;
}

```