#include <bits/stdc++.h>
using namespace std;
const int nMAX = 200005;
const int INF = 0x3f3f3f3f;
class matState {
 public:
  int d[6][6];
  matState(int c = 0) {
    for (int i = 0; i < 6; i++)
      for (int j = 0; j < 6; j++) d[i][j] = c;
  }
  inline matState operator+(const matState &o) const {
    matState r(INF);
    for (int i = 0; i < 6; i++) {
      for (int j = 0; j < 6; j++) {
        for (int k = 0; k < 6; k++) {
          r.d[i][j] = min(r.d[i][j], d[i][k] + o.d[k][j]);
        }
      }
    }
    return r;
  }
};
int n, q;
int t[6][10];
char s[nMAX];
matState tree[4 * nMAX];
void treeBuild(int i = 1, int le = 1, int ri = n) {
  if (le == ri) {
    int x = s[le] - '0';
    tree[i] = matState(INF);
    for (int j = 0; j < 6; j++) {
      tree[i].d[j][t[j][x]] = 0;
      tree[i].d[j][j] = min(tree[i].d[j][j], 1);
    }
  } else {
    int m = (le + ri) / 2;
    treeBuild(2 * i, le, m);
    treeBuild(2 * i + 1, m + 1, ri);
    tree[i] = tree[2 * i] + tree[2 * i + 1];
  }
}
matState treeQuery(int leQ, int riQ, int i = 1, int le = 1, int ri = n) {
  if (leQ <= ri && le <= riQ) {
    if (leQ <= le && ri <= riQ) {
      return tree[i];
    } else {
      int m = (le + ri) / 2;
      return treeQuery(leQ, riQ, i * 2, le, m) +
             treeQuery(leQ, riQ, i * 2 + 1, m + 1, ri);
    }
  } else {
    matState r(INF);
    for (int i = 0; i < 6; i++) r.d[i][i] = 0;
    return r;
  }
}
void fillT() {
  for (int i = 0; i < 6; i++)
    for (int j = 0; j < 10; j++) t[i][j] = i;
  t[0][2] = 1;
  t[1][0] = 2;
  t[2][1] = 3;
  t[3][7] = 4;
  t[3][6] = 5;
  t[4][6] = 5;
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n >> q;
  cin >> (s + 1);
  fillT();
  treeBuild();
  while (q--) {
    int le, ri;
    cin >> le >> ri;
    matState s = treeQuery(le, ri);
    cout << (s.d[0][4] == INF ? -1 : s.d[0][4]) << "\n";
  }
  return 0;
}

### Prompt

Your task is to create a Cpp solution 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;
const int nMAX = 200005;
const int INF = 0x3f3f3f3f;
class matState {
 public:
  int d[6][6];
  matState(int c = 0) {
    for (int i = 0; i < 6; i++)
      for (int j = 0; j < 6; j++) d[i][j] = c;
  }
  inline matState operator+(const matState &o) const {
    matState r(INF);
    for (int i = 0; i < 6; i++) {
      for (int j = 0; j < 6; j++) {
        for (int k = 0; k < 6; k++) {
          r.d[i][j] = min(r.d[i][j], d[i][k] + o.d[k][j]);
        }
      }
    }
    return r;
  }
};
int n, q;
int t[6][10];
char s[nMAX];
matState tree[4 * nMAX];
void treeBuild(int i = 1, int le = 1, int ri = n) {
  if (le == ri) {
    int x = s[le] - '0';
    tree[i] = matState(INF);
    for (int j = 0; j < 6; j++) {
      tree[i].d[j][t[j][x]] = 0;
      tree[i].d[j][j] = min(tree[i].d[j][j], 1);
    }
  } else {
    int m = (le + ri) / 2;
    treeBuild(2 * i, le, m);
    treeBuild(2 * i + 1, m + 1, ri);
    tree[i] = tree[2 * i] + tree[2 * i + 1];
  }
}
matState treeQuery(int leQ, int riQ, int i = 1, int le = 1, int ri = n) {
  if (leQ <= ri && le <= riQ) {
    if (leQ <= le && ri <= riQ) {
      return tree[i];
    } else {
      int m = (le + ri) / 2;
      return treeQuery(leQ, riQ, i * 2, le, m) +
             treeQuery(leQ, riQ, i * 2 + 1, m + 1, ri);
    }
  } else {
    matState r(INF);
    for (int i = 0; i < 6; i++) r.d[i][i] = 0;
    return r;
  }
}
void fillT() {
  for (int i = 0; i < 6; i++)
    for (int j = 0; j < 10; j++) t[i][j] = i;
  t[0][2] = 1;
  t[1][0] = 2;
  t[2][1] = 3;
  t[3][7] = 4;
  t[3][6] = 5;
  t[4][6] = 5;
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n >> q;
  cin >> (s + 1);
  fillT();
  treeBuild();
  while (q--) {
    int le, ri;
    cin >> le >> ri;
    matState s = treeQuery(le, ri);
    cout << (s.d[0][4] == INF ? -1 : s.d[0][4]) << "\n";
  }
  return 0;
}

```