#include <bits/stdc++.h>
using namespace std;
template <typename T>
void out(T x) {
  cout << x << endl;
  exit(0);
}
const int maxn = 2e5 + 200;
const int inf = 1e9;
int n, q;
string s;
struct node {
  int mat[5][5];
  node() {
    for (int i = 0; i < 5; i++) {
      for (int j = 0; j < 5; j++) {
        mat[i][j] = inf;
      }
    }
  }
  node(char c) {
    for (int i = 0; i < 5; i++) {
      for (int j = 0; j < 5; j++) {
        mat[i][j] = i != j ? inf : 0;
      }
    }
    if (c == '2') {
      mat[0][0] = 1;
      mat[0][1] = 0;
    } else if (c == '0') {
      mat[1][1] = 1;
      mat[1][2] = 0;
    } else if (c == '1') {
      mat[2][2] = 1;
      mat[2][3] = 0;
    } else if (c == '7') {
      mat[3][3] = 1;
      mat[3][4] = 0;
    } else if (c == '6') {
      mat[3][3] = 1;
      mat[4][4] = 1;
    } else {
    }
  }
};
node t[4 * maxn];
node merge(node a, node b) {
  node c;
  for (int i = 0; i < 5; i++) {
    for (int j = 0; j < 5; j++) {
      for (int k = 0; k < 5; k++) {
        c.mat[i][j] = min(c.mat[i][j], a.mat[i][k] + b.mat[k][j]);
      }
    }
  }
  return c;
}
void build(int v, int tl, int tr) {
  if (tl == tr) {
    t[v] = node(s[tl]);
  } else {
    int tm = (tl + tr) / 2;
    build(2 * v, tl, tm);
    build(2 * v + 1, tm + 1, tr);
    t[v] = merge(t[2 * v], t[2 * v + 1]);
  }
}
node get(int v, int tl, int tr, int l, int r) {
  if (l == tl && tr == r) {
    return t[v];
  } else {
    int tm = (tl + tr) / 2;
    if (r <= tm) {
      return get(2 * v, tl, tm, l, r);
    } else if (l > tm) {
      return get(2 * v + 1, tm + 1, tr, l, r);
    } else {
      return merge(get(2 * v, tl, tm, l, min(r, tm)),
                   get(2 * v + 1, tm + 1, tr, max(l, tm + 1), r));
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> q;
  cin >> s;
  s = "*" + s;
  build(1, 1, n);
  while (q--) {
    int l, r;
    cin >> l >> r;
    auto res = get(1, 1, n, l, r);
    int ans = res.mat[0][4] == inf ? -1 : res.mat[0][4];
    cout << ans << "\n";
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

A string t is called nice if a string "2017" occurs in t as a subsequence but a string "2016" doesn't occur in t as a subsequence. For example, strings "203434107" and "9220617" are nice, while strings "20016", "1234" and "20167" aren't nice.

The ugliness of a string is the minimum possible number of characters to remove, in order to obtain a nice string. If it's impossible to make a string nice by removing characters, its ugliness is  - 1.

Limak has a string s of length n, with characters indexed 1 through n. He asks you q queries. In the i-th query you should compute and print the ugliness of a substring (continuous subsequence) of s starting at the index ai and ending at the index bi (inclusive).

Input

The first line of the input contains two integers n and q (4 ≤ n ≤ 200 000, 1 ≤ q ≤ 200 000) — the length of the string s and the number of queries respectively.

The second line contains a string s of length n. Every character is one of digits '0'–'9'.

The i-th of next q lines contains two integers ai and bi (1 ≤ ai ≤ bi ≤ n), describing a substring in the i-th query.

Output

For each query print the ugliness of the given substring.

Examples

Input

8 3
20166766
1 8
1 7
2 8


Output

4
3
-1


Input

15 5
012016662091670
3 4
1 14
4 15
1 13
10 15


Output

-1
2
1
-1
-1


Input

4 2
1234
2 4
1 2


Output

-1
-1

Note

In the first sample:

  * In the first query, ugliness("20166766") = 4 because all four sixes must be removed. 
  * In the second query, ugliness("2016676") = 3 because all three sixes must be removed. 
  * In the third query, ugliness("0166766") = - 1 because it's impossible to remove some digits to get a nice string. 



In the second sample:

  * In the second query, ugliness("01201666209167") = 2. It's optimal to remove the first digit '2' and the last digit '6', what gives a string "010166620917", which is nice. 
  * In the third query, ugliness("016662091670") = 1. It's optimal to remove the last digit '6', what gives a nice string "01666209170". 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <typename T>
void out(T x) {
  cout << x << endl;
  exit(0);
}
const int maxn = 2e5 + 200;
const int inf = 1e9;
int n, q;
string s;
struct node {
  int mat[5][5];
  node() {
    for (int i = 0; i < 5; i++) {
      for (int j = 0; j < 5; j++) {
        mat[i][j] = inf;
      }
    }
  }
  node(char c) {
    for (int i = 0; i < 5; i++) {
      for (int j = 0; j < 5; j++) {
        mat[i][j] = i != j ? inf : 0;
      }
    }
    if (c == '2') {
      mat[0][0] = 1;
      mat[0][1] = 0;
    } else if (c == '0') {
      mat[1][1] = 1;
      mat[1][2] = 0;
    } else if (c == '1') {
      mat[2][2] = 1;
      mat[2][3] = 0;
    } else if (c == '7') {
      mat[3][3] = 1;
      mat[3][4] = 0;
    } else if (c == '6') {
      mat[3][3] = 1;
      mat[4][4] = 1;
    } else {
    }
  }
};
node t[4 * maxn];
node merge(node a, node b) {
  node c;
  for (int i = 0; i < 5; i++) {
    for (int j = 0; j < 5; j++) {
      for (int k = 0; k < 5; k++) {
        c.mat[i][j] = min(c.mat[i][j], a.mat[i][k] + b.mat[k][j]);
      }
    }
  }
  return c;
}
void build(int v, int tl, int tr) {
  if (tl == tr) {
    t[v] = node(s[tl]);
  } else {
    int tm = (tl + tr) / 2;
    build(2 * v, tl, tm);
    build(2 * v + 1, tm + 1, tr);
    t[v] = merge(t[2 * v], t[2 * v + 1]);
  }
}
node get(int v, int tl, int tr, int l, int r) {
  if (l == tl && tr == r) {
    return t[v];
  } else {
    int tm = (tl + tr) / 2;
    if (r <= tm) {
      return get(2 * v, tl, tm, l, r);
    } else if (l > tm) {
      return get(2 * v + 1, tm + 1, tr, l, r);
    } else {
      return merge(get(2 * v, tl, tm, l, min(r, tm)),
                   get(2 * v + 1, tm + 1, tr, max(l, tm + 1), r));
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> q;
  cin >> s;
  s = "*" + s;
  build(1, 1, n);
  while (q--) {
    int l, r;
    cin >> l >> r;
    auto res = get(1, 1, n, l, r);
    int ans = res.mat[0][4] == inf ? -1 : res.mat[0][4];
    cout << ans << "\n";
  }
  return 0;
}

```