#include <bits/stdc++.h>
using namespace std;
const string good = "2017";
const string bad = "2016";
struct Data {
  int cost[5][5];
  Data() {
    for (int i = 0; i < 5; i++) {
      for (int j = 0; j < 5; j++) {
        cost[i][j] = 1e9;
      }
    }
    for (int i = 0; i < 5; i++) cost[i][i] = 0;
  }
  Data operator+(Data o) {
    Data res;
    for (int i = 0; i < 5; i++) res.cost[i][i] = 1e9;
    for (int i = 0; i < 5; i++) {
      for (int j = i; j < 5; j++) {
        for (int k = j; k < 5; k++) {
          res.cost[i][k] = min(res.cost[i][k], cost[i][j] + o.cost[j][k]);
        }
      }
    }
    return res;
  }
};
char s[200013];
class ST {
 private:
  int size;
  Data st[1 << 19];
  void build(int w, int L, int R) {
    if (L == R) {
      if (s[L] == bad[3]) {
        st[w].cost[3][3] = 1;
        st[w].cost[4][4] = 1;
      }
      for (int i = 0; i < 4; i++) {
        if (s[L] == good[i]) {
          st[w].cost[i][i + 1] = 0;
          st[w].cost[i][i] = 1;
        }
      }
    } else {
      build(w * 2, L, (L + R) / 2), build(w * 2 + 1, (L + R) / 2 + 1, R);
      st[w] = st[w * 2] + st[w * 2 + 1];
    }
  }
  Data query(int w, int L, int R, int a, int b) {
    if (b < L || R < a) return Data();
    if (a <= L && R <= b) return st[w];
    return query(w * 2, L, (L + R) / 2, a, b) +
           query(w * 2 + 1, (L + R) / 2 + 1, R, a, b);
  }

 public:
  void build(int n) {
    size = n;
    build(1, 1, size);
  }
  Data query(int a, int b) { return query(1, 1, size, a, b); }
};
int n, q;
ST tree;
int main() {
  scanf("%d%d", &n, &q);
  scanf(" %s", s + 1);
  tree.build(n);
  while (q--) {
    int a, b;
    scanf("%d%d", &a, &b);
    int ans = tree.query(a, b).cost[0][4];
    if (ans <= n)
      printf("%d\n", ans);
    else
      printf("-1\n");
  }
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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 string good = "2017";
const string bad = "2016";
struct Data {
  int cost[5][5];
  Data() {
    for (int i = 0; i < 5; i++) {
      for (int j = 0; j < 5; j++) {
        cost[i][j] = 1e9;
      }
    }
    for (int i = 0; i < 5; i++) cost[i][i] = 0;
  }
  Data operator+(Data o) {
    Data res;
    for (int i = 0; i < 5; i++) res.cost[i][i] = 1e9;
    for (int i = 0; i < 5; i++) {
      for (int j = i; j < 5; j++) {
        for (int k = j; k < 5; k++) {
          res.cost[i][k] = min(res.cost[i][k], cost[i][j] + o.cost[j][k]);
        }
      }
    }
    return res;
  }
};
char s[200013];
class ST {
 private:
  int size;
  Data st[1 << 19];
  void build(int w, int L, int R) {
    if (L == R) {
      if (s[L] == bad[3]) {
        st[w].cost[3][3] = 1;
        st[w].cost[4][4] = 1;
      }
      for (int i = 0; i < 4; i++) {
        if (s[L] == good[i]) {
          st[w].cost[i][i + 1] = 0;
          st[w].cost[i][i] = 1;
        }
      }
    } else {
      build(w * 2, L, (L + R) / 2), build(w * 2 + 1, (L + R) / 2 + 1, R);
      st[w] = st[w * 2] + st[w * 2 + 1];
    }
  }
  Data query(int w, int L, int R, int a, int b) {
    if (b < L || R < a) return Data();
    if (a <= L && R <= b) return st[w];
    return query(w * 2, L, (L + R) / 2, a, b) +
           query(w * 2 + 1, (L + R) / 2 + 1, R, a, b);
  }

 public:
  void build(int n) {
    size = n;
    build(1, 1, size);
  }
  Data query(int a, int b) { return query(1, 1, size, a, b); }
};
int n, q;
ST tree;
int main() {
  scanf("%d%d", &n, &q);
  scanf(" %s", s + 1);
  tree.build(n);
  while (q--) {
    int a, b;
    scanf("%d%d", &a, &b);
    int ans = tree.query(a, b).cost[0][4];
    if (ans <= n)
      printf("%d\n", ans);
    else
      printf("-1\n");
  }
  return 0;
}

```