Problem: JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.

First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.

For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.

However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.

All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.

Input

The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).

The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.

Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.

Output

Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.

Examples

Input

4 2
1011
1 4
3 4


Output

14
3


Input

3 2
111
1 2
3 3


Output

3
1

Note

In the first example: 

  * For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2. 
  * For query 2: Both 3, 4 and 4, 3 ordering give the same answer. 



In the second example, any order of eating parts leads to the same answer.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int inf = 1000000000;
const long long infLL = 1000000000000000000;
long long MOD = 1000000000;
inline bool checkbit(long long n, int i) { return n & (1LL << i); }
inline long long setbit(long long n, int i) { return n | (1LL << i); }
inline long long resetbit(long long n, int i) { return n & (~(1LL << i)); }
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
inline bool isLeapYear(long long year) {
  return (year % 400 == 0) || (year % 4 == 0 && year % 100 != 0);
}
inline bool isInside(pair<int, int> p, long long n, long long m) {
  return (p.first >= 0 && p.first < n && p.second >= 0 && p.second < m);
}
inline bool isInside(pair<int, int> p, long long n) {
  return (p.first >= 0 && p.first < n && p.second >= 0 && p.second <= n);
}
inline bool isSquare(long long x) {
  long long s = sqrt(x);
  return (s * s == x);
}
inline bool isFib(long long x) {
  return isSquare(5 * x * x + 4) || isSquare(5 * x * x - 4);
}
inline bool isPowerOfTwo(long long x) { return (x && !(x & (x - 1))); }
inline void normal(long long &a, long long MOD) {
  a %= MOD;
  (a < 0) && (a += MOD);
}
inline long long modMul(long long a, long long b, long long MOD) {
  a %= MOD, b %= MOD;
  normal(a, MOD), normal(b, MOD);
  return (a * b) % MOD;
}
inline long long modAdd(long long a, long long b, long long MOD) {
  a %= MOD, b %= MOD;
  normal(a, MOD), normal(b, MOD);
  return (a + b) % MOD;
}
inline long long modSub(long long a, long long b, long long MOD) {
  a %= MOD, b %= MOD;
  normal(a, MOD), normal(b, MOD);
  a -= b;
  normal(a, MOD);
  return a;
}
inline long long modPow(long long b, long long p, long long MOD) {
  long long r = 1;
  while (p) {
    if (p & 1) r = modMul(r, b, MOD);
    b = modMul(b, b, MOD);
    p >>= 1;
  }
  return r;
}
inline long long modInverse(long long a, long long MOD) {
  return modPow(a, MOD - 2, MOD);
}
inline long long modDiv(long long a, long long b, long long MOD) {
  return modMul(a, modInverse(b, MOD), MOD);
}
struct func {
  bool operator()(pair<int, int> const &a, pair<int, int> const &b) {
    if (a.first == b.first) {
      return (a.second < b.second);
    }
    return (a.first < b.first);
  }
};
long long a[100005], one[100005], zero[100005], pw[100005], oo, zz, n, q, x, y;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  string s;
  cin >> n >> q >> s;
  for (int i = 1; i <= n; ++i) {
    a[i] = s[i - 1] - '0';
    if (a[i] == 1LL) {
      one[i] = 1;
    } else {
      zero[i] = 1;
    }
    one[i] += one[i - 1];
    zero[i] += zero[i - 1];
  }
  pw[0] = 1LL;
  for (int i = 1; i <= 100005; ++i) {
    pw[i] = (pw[i - 1] * 2LL) % 1000000007;
  }
  for (int i = 1; i <= q; ++i) {
    cin >> x >> y;
    oo = one[y] - one[x - 1];
    zz = zero[y] - zero[x - 1];
    long long ooo = pw[oo] - 1;
    long long zzz = pw[zz];
    long long ans = modMul(ooo, zzz, 1000000007);
    cout << ans << '\n';
  }
  return 0;
}