#include <bits/stdc++.h>
using namespace std;
const int maxn = 200010;
const int inf = 0x3f3f3f3f;
int n, m;
char str[maxn];
struct Node {
  int f[5][5];
  Node() { memset(f, 0x3f, sizeof(f)); }
  void Clear() { memset(f, 0x3f, sizeof(f)); }
} T[maxn << 2];
Node operator+(const Node &a, const Node &b) {
  Node c;
  for (int i = 0; i < 5; ++i) {
    for (int j = i; j < 5; ++j) {
      for (int k = i; k <= j; ++k) {
        c.f[i][j] = min(c.f[i][j], a.f[i][k] + b.f[k][j]);
      }
    }
  }
  return c;
}
void Build(int rt, int l, int r) {
  if (l == r) {
    for (int i = 0; i < 5; ++i) T[rt].f[i][i] = 0;
    if (str[l] == '2')
      T[rt].f[0][1] = 0, T[rt].f[0][0] = 1;
    else if (str[l] == '0')
      T[rt].f[1][2] = 0, T[rt].f[1][1] = 1;
    else if (str[l] == '1')
      T[rt].f[2][3] = 0, T[rt].f[2][2] = 1;
    else if (str[l] == '7')
      T[rt].f[3][4] = 0, T[rt].f[3][3] = 1;
    else if (str[l] == '6')
      T[rt].f[3][3] = T[rt].f[4][4] = 1;
    return;
  }
  Build((rt << 1), l, ((l + r) >> 1));
  Build((rt << 1 | 1), ((l + r) >> 1) + 1, r);
  T[rt] = T[(rt << 1)] + T[(rt << 1 | 1)];
}
int s, t;
Node res;
void Query(int rt, int l, int r) {
  if (s <= l && t >= r) {
    res = res + T[rt];
    return;
  }
  if (s <= ((l + r) >> 1)) Query((rt << 1), l, ((l + r) >> 1));
  if (t > ((l + r) >> 1)) Query((rt << 1 | 1), ((l + r) >> 1) + 1, r);
}
int main() {
  scanf("%d%d", &n, &m);
  scanf("%s", str + 1);
  Build(1, 1, n);
  while (m--) {
    scanf("%d%d", &s, &t);
    res.Clear();
    res.f[0][0] = 0;
    Query(1, 1, n);
    if (res.f[0][4] >= inf)
      puts("-1");
    else
      printf("%d\n", res.f[0][4]);
  }
  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;
const int maxn = 200010;
const int inf = 0x3f3f3f3f;
int n, m;
char str[maxn];
struct Node {
  int f[5][5];
  Node() { memset(f, 0x3f, sizeof(f)); }
  void Clear() { memset(f, 0x3f, sizeof(f)); }
} T[maxn << 2];
Node operator+(const Node &a, const Node &b) {
  Node c;
  for (int i = 0; i < 5; ++i) {
    for (int j = i; j < 5; ++j) {
      for (int k = i; k <= j; ++k) {
        c.f[i][j] = min(c.f[i][j], a.f[i][k] + b.f[k][j]);
      }
    }
  }
  return c;
}
void Build(int rt, int l, int r) {
  if (l == r) {
    for (int i = 0; i < 5; ++i) T[rt].f[i][i] = 0;
    if (str[l] == '2')
      T[rt].f[0][1] = 0, T[rt].f[0][0] = 1;
    else if (str[l] == '0')
      T[rt].f[1][2] = 0, T[rt].f[1][1] = 1;
    else if (str[l] == '1')
      T[rt].f[2][3] = 0, T[rt].f[2][2] = 1;
    else if (str[l] == '7')
      T[rt].f[3][4] = 0, T[rt].f[3][3] = 1;
    else if (str[l] == '6')
      T[rt].f[3][3] = T[rt].f[4][4] = 1;
    return;
  }
  Build((rt << 1), l, ((l + r) >> 1));
  Build((rt << 1 | 1), ((l + r) >> 1) + 1, r);
  T[rt] = T[(rt << 1)] + T[(rt << 1 | 1)];
}
int s, t;
Node res;
void Query(int rt, int l, int r) {
  if (s <= l && t >= r) {
    res = res + T[rt];
    return;
  }
  if (s <= ((l + r) >> 1)) Query((rt << 1), l, ((l + r) >> 1));
  if (t > ((l + r) >> 1)) Query((rt << 1 | 1), ((l + r) >> 1) + 1, r);
}
int main() {
  scanf("%d%d", &n, &m);
  scanf("%s", str + 1);
  Build(1, 1, n);
  while (m--) {
    scanf("%d%d", &s, &t);
    res.Clear();
    res.f[0][0] = 0;
    Query(1, 1, n);
    if (res.f[0][4] >= inf)
      puts("-1");
    else
      printf("%d\n", res.f[0][4]);
  }
  return 0;
}

```