#include <bits/stdc++.h>
using namespace std;
const int Maxn = 200005;
const int Maxm = 1048576;
const int Maxd = 5;
const int Inf = 1000000000;
struct node {
  int dp[Maxd][Maxd];
  node(int my = -1) {
    for (int i = 0; i < Maxd; i++) {
      for (int j = i + 1; j < Maxd; j++) dp[i][j] = Inf;
      if (my == 2)
        if (i == 0) {
          dp[i][i + 1] = 0;
          dp[i][i] = 1;
        } else
          dp[i][i] = 0;
      else if (my == 0)
        if (i == 1) {
          dp[i][i + 1] = 0;
          dp[i][i] = 1;
        } else
          dp[i][i] = 0;
      else if (my == 1)
        if (i == 2) {
          dp[i][i + 1] = 0;
          dp[i][i] = 1;
        } else
          dp[i][i] = 0;
      else if (my == 7)
        if (i == 3) {
          dp[i][i + 1] = 0;
          dp[i][i] = 1;
        } else
          dp[i][i] = 0;
      else if (my == 6)
        if (i < 3)
          dp[i][i] = 0;
        else
          dp[i][i] = 1;
      else
        dp[i][i] = my == -1 ? Inf : 0;
    }
  }
  node operator+(const node &b) const {
    node c;
    for (int i = 0; i < Maxd; i++)
      for (int j = i; j < Maxd; j++)
        for (int k = j; k < Maxd; k++)
          c.dp[i][k] = min(c.dp[i][k], dp[i][j] + b.dp[j][k]);
    return c;
  }
};
int n, q;
char str[Maxn];
node st[Maxm];
void Create(int v, int l, int r) {
  if (l == r)
    st[v] = node(str[l] - '0');
  else {
    int m = l + r >> 1;
    Create(2 * v, l, m);
    Create(2 * v + 1, m + 1, r);
    st[v] = st[2 * v] + st[2 * v + 1];
  }
}
node Get(int v, int l, int r, int a, int b) {
  if (l == a && r == b)
    return st[v];
  else {
    int m = l + r >> 1;
    if (b <= m) return Get(2 * v, l, m, a, b);
    if (m + 1 <= a) return Get(2 * v + 1, m + 1, r, a, b);
    return Get(2 * v, l, m, a, m) + Get(2 * v + 1, m + 1, r, m + 1, b);
  }
}
int main() {
  scanf("%d %d", &n, &q);
  scanf("%s", str + 1);
  Create(1, 1, n);
  while (q--) {
    int l, r;
    scanf("%d %d", &l, &r);
    int res = Get(1, 1, n, l, r).dp[0][4];
    if (res >= Inf) res = -1;
    printf("%d\n", res);
  }
  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 int Maxn = 200005;
const int Maxm = 1048576;
const int Maxd = 5;
const int Inf = 1000000000;
struct node {
  int dp[Maxd][Maxd];
  node(int my = -1) {
    for (int i = 0; i < Maxd; i++) {
      for (int j = i + 1; j < Maxd; j++) dp[i][j] = Inf;
      if (my == 2)
        if (i == 0) {
          dp[i][i + 1] = 0;
          dp[i][i] = 1;
        } else
          dp[i][i] = 0;
      else if (my == 0)
        if (i == 1) {
          dp[i][i + 1] = 0;
          dp[i][i] = 1;
        } else
          dp[i][i] = 0;
      else if (my == 1)
        if (i == 2) {
          dp[i][i + 1] = 0;
          dp[i][i] = 1;
        } else
          dp[i][i] = 0;
      else if (my == 7)
        if (i == 3) {
          dp[i][i + 1] = 0;
          dp[i][i] = 1;
        } else
          dp[i][i] = 0;
      else if (my == 6)
        if (i < 3)
          dp[i][i] = 0;
        else
          dp[i][i] = 1;
      else
        dp[i][i] = my == -1 ? Inf : 0;
    }
  }
  node operator+(const node &b) const {
    node c;
    for (int i = 0; i < Maxd; i++)
      for (int j = i; j < Maxd; j++)
        for (int k = j; k < Maxd; k++)
          c.dp[i][k] = min(c.dp[i][k], dp[i][j] + b.dp[j][k]);
    return c;
  }
};
int n, q;
char str[Maxn];
node st[Maxm];
void Create(int v, int l, int r) {
  if (l == r)
    st[v] = node(str[l] - '0');
  else {
    int m = l + r >> 1;
    Create(2 * v, l, m);
    Create(2 * v + 1, m + 1, r);
    st[v] = st[2 * v] + st[2 * v + 1];
  }
}
node Get(int v, int l, int r, int a, int b) {
  if (l == a && r == b)
    return st[v];
  else {
    int m = l + r >> 1;
    if (b <= m) return Get(2 * v, l, m, a, b);
    if (m + 1 <= a) return Get(2 * v + 1, m + 1, r, a, b);
    return Get(2 * v, l, m, a, m) + Get(2 * v + 1, m + 1, r, m + 1, b);
  }
}
int main() {
  scanf("%d %d", &n, &q);
  scanf("%s", str + 1);
  Create(1, 1, n);
  while (q--) {
    int l, r;
    scanf("%d %d", &l, &r);
    int res = Get(1, 1, n, l, r).dp[0][4];
    if (res >= Inf) res = -1;
    printf("%d\n", res);
  }
  return 0;
}

```