#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
struct rec {
  int dp[5][5];
  void init() {
    for (int i = 0; i < 5; ++i)
      for (int j = i; j < 5; ++j)
        if (i == j)
          dp[i][j] = 0;
        else
          dp[i][j] = inf;
  }
  rec operator+(const rec &a) const {
    rec b;
    for (int i = 0; i < 5; ++i)
      for (int j = i; j < 5; ++j) {
        b.dp[i][j] = inf;
        for (int k = i; k <= j; ++k)
          b.dp[i][j] = min(b.dp[i][j], dp[i][k] + a.dp[k][j]);
      }
    return b;
  }
} seg[200010 * 3], ans;
char st[200010];
int n, m;
void build(int x, int s, int t) {
  if (s == t) {
    seg[x].init();
    int i = s - 1;
    if (st[i] == '2')
      seg[x].dp[0][0] = 1, seg[x].dp[0][1] = 0;
    else if (st[i] == '0')
      seg[x].dp[1][1] = 1, seg[x].dp[1][2] = 0;
    else if (st[i] == '1')
      seg[x].dp[2][2] = 1, seg[x].dp[2][3] = 0;
    else if (st[i] == '7')
      seg[x].dp[3][3] = 1, seg[x].dp[3][4] = 0;
    else if (st[i] == '6')
      seg[x].dp[3][3] = 1, seg[x].dp[4][4] = 1;
    return;
  }
  int mid = s + t >> 1;
  build(x << 1, s, mid);
  build(x << 1 | 1, mid + 1, t);
  seg[x] = seg[x << 1] + seg[x << 1 | 1];
}
void query(int x, int s, int t, int l, int r) {
  if (l <= s && t <= r) {
    ans = ans + seg[x];
    return;
  }
  int mid = s + t >> 1;
  if (l <= mid) query(x << 1, s, mid, l, r);
  if (mid < r) query(x << 1 | 1, mid + 1, t, l, r);
}
int main() {
  scanf("%d%d", &n, &m);
  scanf("%s", st);
  build(1, 1, n);
  int l, r;
  while (m--) {
    scanf("%d%d", &l, &r);
    ans.init();
    query(1, 1, n, l, r);
    if (ans.dp[0][4] == inf)
      printf("-1\n");
    else
      printf("%d\n", ans.dp[0][4]);
  }
  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 inf = 0x3f3f3f3f;
struct rec {
  int dp[5][5];
  void init() {
    for (int i = 0; i < 5; ++i)
      for (int j = i; j < 5; ++j)
        if (i == j)
          dp[i][j] = 0;
        else
          dp[i][j] = inf;
  }
  rec operator+(const rec &a) const {
    rec b;
    for (int i = 0; i < 5; ++i)
      for (int j = i; j < 5; ++j) {
        b.dp[i][j] = inf;
        for (int k = i; k <= j; ++k)
          b.dp[i][j] = min(b.dp[i][j], dp[i][k] + a.dp[k][j]);
      }
    return b;
  }
} seg[200010 * 3], ans;
char st[200010];
int n, m;
void build(int x, int s, int t) {
  if (s == t) {
    seg[x].init();
    int i = s - 1;
    if (st[i] == '2')
      seg[x].dp[0][0] = 1, seg[x].dp[0][1] = 0;
    else if (st[i] == '0')
      seg[x].dp[1][1] = 1, seg[x].dp[1][2] = 0;
    else if (st[i] == '1')
      seg[x].dp[2][2] = 1, seg[x].dp[2][3] = 0;
    else if (st[i] == '7')
      seg[x].dp[3][3] = 1, seg[x].dp[3][4] = 0;
    else if (st[i] == '6')
      seg[x].dp[3][3] = 1, seg[x].dp[4][4] = 1;
    return;
  }
  int mid = s + t >> 1;
  build(x << 1, s, mid);
  build(x << 1 | 1, mid + 1, t);
  seg[x] = seg[x << 1] + seg[x << 1 | 1];
}
void query(int x, int s, int t, int l, int r) {
  if (l <= s && t <= r) {
    ans = ans + seg[x];
    return;
  }
  int mid = s + t >> 1;
  if (l <= mid) query(x << 1, s, mid, l, r);
  if (mid < r) query(x << 1 | 1, mid + 1, t, l, r);
}
int main() {
  scanf("%d%d", &n, &m);
  scanf("%s", st);
  build(1, 1, n);
  int l, r;
  while (m--) {
    scanf("%d%d", &l, &r);
    ans.init();
    query(1, 1, n, l, r);
    if (ans.dp[0][4] == inf)
      printf("-1\n");
    else
      printf("%d\n", ans.dp[0][4]);
  }
  return 0;
}

```