#include <bits/stdc++.h>
using namespace std;
struct E {
  int mat[5][5];
};
const int inf = 1e8;
E null;
E reset(char a) {
  E w;
  for (int i = 0; i < 5; i++) {
    for (int j = 0; j < 5; j++) {
      w.mat[i][j] = inf;
    }
    w.mat[i][i] = 0;
  }
  if (a == '2') {
    w.mat[0][0] = 1;
    w.mat[0][1] = 0;
  }
  if (a == '0') {
    w.mat[1][1] = 1;
    w.mat[1][2] = 0;
  }
  if (a == '1') {
    w.mat[2][2] = 1;
    w.mat[2][3] = 0;
  }
  if (a == '6') {
    w.mat[3][3] = 1;
    w.mat[4][4] = 1;
  }
  if (a == '7') {
    w.mat[3][4] = 0;
    w.mat[3][3] = 1;
  }
  return w;
}
E combine(E a, E b) {
  E w;
  for (int i = 0; i < 5; i++) {
    for (int j = 0; j < 5; j++) {
      w.mat[i][j] = inf;
    }
  }
  for (int i = 0; i < 5; i++) {
    for (int j = 0; j < 5; j++) {
      for (int k = 0; k < 5; k++) {
        w.mat[i][j] = min(w.mat[i][j], a.mat[i][k] + b.mat[k][j]);
      }
    }
  }
  return w;
}
string second;
E tree[800011];
struct node {
  int l, r, id;
  node(){};
  node(int _l, int _r, int _id) {
    l = _l;
    r = _r;
    id = _id;
  }
  node left() { return node(l, (l + r) / 2, 2 * id); }
  node right() { return node((l + r) / 2 + 1, r, 2 * id + 1); }
  void build() {
    if (l == r) {
      tree[id] = reset(second[l]);
      return;
    }
    left().build();
    right().build();
    tree[id] = combine(tree[2 * id], tree[2 * id + 1]);
  }
  E query(int second, int e) {
    if (r < second or e < l) {
      return null;
    }
    if (second <= l and r <= e) {
      return tree[id];
    }
    return combine(left().query(second, e), right().query(second, e));
  }
};
int main() {
  for (int i = 0; i < 5; i++) {
    for (int j = 0; j < 5; j++) {
      null.mat[i][j] = inf;
    }
    null.mat[i][i] = 0;
  }
  int n, q, l, r;
  cin >> n >> q;
  cin >> second;
  second = '$' + second;
  node stree(1, n, 1);
  stree.build();
  while (q--) {
    cin >> l >> r;
    int ret = stree.query(l, r).mat[0][4];
    if (ret == inf) ret = -1;
    cout << ret << "\n";
  }
}

### Prompt

Please create a solution in CPP to the following problem:

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;
struct E {
  int mat[5][5];
};
const int inf = 1e8;
E null;
E reset(char a) {
  E w;
  for (int i = 0; i < 5; i++) {
    for (int j = 0; j < 5; j++) {
      w.mat[i][j] = inf;
    }
    w.mat[i][i] = 0;
  }
  if (a == '2') {
    w.mat[0][0] = 1;
    w.mat[0][1] = 0;
  }
  if (a == '0') {
    w.mat[1][1] = 1;
    w.mat[1][2] = 0;
  }
  if (a == '1') {
    w.mat[2][2] = 1;
    w.mat[2][3] = 0;
  }
  if (a == '6') {
    w.mat[3][3] = 1;
    w.mat[4][4] = 1;
  }
  if (a == '7') {
    w.mat[3][4] = 0;
    w.mat[3][3] = 1;
  }
  return w;
}
E combine(E a, E b) {
  E w;
  for (int i = 0; i < 5; i++) {
    for (int j = 0; j < 5; j++) {
      w.mat[i][j] = inf;
    }
  }
  for (int i = 0; i < 5; i++) {
    for (int j = 0; j < 5; j++) {
      for (int k = 0; k < 5; k++) {
        w.mat[i][j] = min(w.mat[i][j], a.mat[i][k] + b.mat[k][j]);
      }
    }
  }
  return w;
}
string second;
E tree[800011];
struct node {
  int l, r, id;
  node(){};
  node(int _l, int _r, int _id) {
    l = _l;
    r = _r;
    id = _id;
  }
  node left() { return node(l, (l + r) / 2, 2 * id); }
  node right() { return node((l + r) / 2 + 1, r, 2 * id + 1); }
  void build() {
    if (l == r) {
      tree[id] = reset(second[l]);
      return;
    }
    left().build();
    right().build();
    tree[id] = combine(tree[2 * id], tree[2 * id + 1]);
  }
  E query(int second, int e) {
    if (r < second or e < l) {
      return null;
    }
    if (second <= l and r <= e) {
      return tree[id];
    }
    return combine(left().query(second, e), right().query(second, e));
  }
};
int main() {
  for (int i = 0; i < 5; i++) {
    for (int j = 0; j < 5; j++) {
      null.mat[i][j] = inf;
    }
    null.mat[i][i] = 0;
  }
  int n, q, l, r;
  cin >> n >> q;
  cin >> second;
  second = '$' + second;
  node stree(1, n, 1);
  stree.build();
  while (q--) {
    cin >> l >> r;
    int ret = stree.query(l, r).mat[0][4];
    if (ret == inf) ret = -1;
    cout << ret << "\n";
  }
}

```