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;
vector<long long> _data;
long long a, q, b, k, l, r, n, m, tmp, ans = 0;
template <int m>
struct modint {
  long long x;
  modint() : x(0) {}
  modint(long long arg) {
    arg %= m;
    if (arg < 0)
      x = arg + m;
    else
      x = arg;
  }
  modint& operator+=(const modint& other) {
    x += other.x;
    if (x >= m) x -= m;
    return *this;
  }
  modint& operator*=(const modint& other) {
    x = (x * 1ll * other.x) % m;
    return *this;
  }
  modint& operator-=(const modint& other) {
    x += m - other.x;
    if (x >= m) x -= m;
    return *this;
  }
  modint operator+(const modint& other) const {
    modint tmp = *this;
    tmp += other;
    return tmp;
  }
  modint operator-(const modint& other) const {
    modint tmp = *this;
    tmp -= other;
    return tmp;
  }
  modint operator*(const modint& other) const {
    modint tmp = *this;
    tmp *= other;
    return tmp;
  }
  explicit operator int() const { return x; }
  modint& operator++() {
    ++x;
    if (x == m) x = 0;
    return *this;
  }
  modint& operator--() {
    if (x == 0)
      x = m - 1;
    else
      --x;
    return *this;
  }
  modint operator++(int) {
    modint tmp = *this;
    ++*this;
    return tmp;
  }
  modint operator--(int) {
    modint tmp = *this;
    --*this;
    return tmp;
  }
  bool operator==(const modint& other) const { return x == other.x; }
  bool operator!=(const modint& other) const { return x != other.x; }
  template <class T>
  modint operator^(T arg) const {
    if (arg == 0) return 1;
    if (arg == 1) return x;
    auto t = *this ^ (arg >> 1);
    t *= t;
    if (arg & 1) t *= *this;
    return t;
  }
  modint inv() const { return *this ^ (m - 2); }
};
const int MOD = 1000000007;
int get_ans(int x, int y) {
  if (x == 0) return 0;
  modint<MOD> ansx(2);
  ansx = ansx ^ (x);
  ansx--;
  if (y == 0) {
    return ansx.x;
  }
  modint<MOD> ansy(2);
  ansy = ansy ^ (y);
  ansy--;
  ansy *= ansx;
  ansy += ansx;
  return ansy.x;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cin >> n >> q;
  string text;
  cin >> text;
  _data.resize(n + 1);
  _data[0] = 0;
  for (int i = 1; i <= n; i++) {
    _data[i] = _data[i - 1] + (text[i - 1] - '0');
  }
  for (int i = 0; i < q; i++) {
    cin >> l >> r;
    --l;
    int num = _data[r] - _data[l];
    int dist = r - l;
    cout << get_ans(num, dist - num) << "\n";
  }
  return 0;
}