#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline bool smin(T &a, const T &b) {
  return b < a ? a = b, 1 : 0;
}
template <typename T>
inline bool smax(T &a, const T &b) {
  return a < b ? a = b, 1 : 0;
}
const long long N = (long long)2e5 + 10;
long long n, q;
string s;
struct node {
  long long d[5][5];
  node() { memset(d, 63, sizeof d); }
  node operator+(node other) {
    node ret;
    for (long long k = 0; k < 5; ++k) {
      for (long long i = 0; i < 5; ++i) {
        for (long long j = 0; j < 5; ++j) {
          smin(ret.d[i][j], d[i][k] + other.d[k][j]);
        }
      }
    }
    return ret;
  }
} seg[N << 2];
void do_erase(long long id, long long v, long long u = -1) {
  for (long long i = 0; i < 5; ++i) {
    if (i == v || i == u) continue;
    seg[id].d[i][i] = 0;
  }
}
void build(long long id = 1, long long l = 0, long long r = n) {
  if (r - l < 2) {
    if (s[l] == '2') {
      seg[id].d[0][0] = 1;
      seg[id].d[0][1] = 0;
      do_erase(id, 0);
    } else if (s[l] == '0') {
      seg[id].d[1][1] = 1;
      seg[id].d[1][2] = 0;
      do_erase(id, 1);
    } else if (s[l] == '1') {
      seg[id].d[2][2] = 1;
      seg[id].d[2][3] = 0;
      do_erase(id, 2);
    } else if (s[l] == '6') {
      seg[id].d[3][3] = 1;
      seg[id].d[4][4] = 1;
      do_erase(id, 3, 4);
    } else if (s[l] == '7') {
      seg[id].d[3][4] = 0;
      seg[id].d[3][3] = 1;
      do_erase(id, 3);
    } else {
      for (long long i = 0; i < 5; ++i) seg[id].d[i][i] = 0;
    }
    return;
  }
  long long mid = (r + l) >> 1, lc = id << 1, rc = lc | 1;
  build(lc, l, mid);
  build(rc, mid, r);
  seg[id] = seg[lc] + seg[rc];
}
node get(long long x, long long y, long long id = 1, long long l = 0,
         long long r = n) {
  if (r <= x || l >= y) {
    node hichi;
    for (long long i = 0; i < 5; ++i) hichi.d[i][i] = 0;
    return hichi;
  }
  if (l >= x && r <= y) return seg[id];
  long long mid = (r + l) >> 1, lc = id << 1, rc = lc | 1;
  return (get(x, y, lc, l, mid) + get(x, y, rc, mid, r));
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> q >> s;
  build();
  while (q--) {
    long long l, r;
    cin >> l >> r;
    node res = get(--l, r);
    if (res.d[0][4] > n) res.d[0][4] = -1;
    cout << res.d[0][4] << endl;
  }
}

### Prompt

Please provide a CPP coded solution 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;
template <typename T>
inline bool smin(T &a, const T &b) {
  return b < a ? a = b, 1 : 0;
}
template <typename T>
inline bool smax(T &a, const T &b) {
  return a < b ? a = b, 1 : 0;
}
const long long N = (long long)2e5 + 10;
long long n, q;
string s;
struct node {
  long long d[5][5];
  node() { memset(d, 63, sizeof d); }
  node operator+(node other) {
    node ret;
    for (long long k = 0; k < 5; ++k) {
      for (long long i = 0; i < 5; ++i) {
        for (long long j = 0; j < 5; ++j) {
          smin(ret.d[i][j], d[i][k] + other.d[k][j]);
        }
      }
    }
    return ret;
  }
} seg[N << 2];
void do_erase(long long id, long long v, long long u = -1) {
  for (long long i = 0; i < 5; ++i) {
    if (i == v || i == u) continue;
    seg[id].d[i][i] = 0;
  }
}
void build(long long id = 1, long long l = 0, long long r = n) {
  if (r - l < 2) {
    if (s[l] == '2') {
      seg[id].d[0][0] = 1;
      seg[id].d[0][1] = 0;
      do_erase(id, 0);
    } else if (s[l] == '0') {
      seg[id].d[1][1] = 1;
      seg[id].d[1][2] = 0;
      do_erase(id, 1);
    } else if (s[l] == '1') {
      seg[id].d[2][2] = 1;
      seg[id].d[2][3] = 0;
      do_erase(id, 2);
    } else if (s[l] == '6') {
      seg[id].d[3][3] = 1;
      seg[id].d[4][4] = 1;
      do_erase(id, 3, 4);
    } else if (s[l] == '7') {
      seg[id].d[3][4] = 0;
      seg[id].d[3][3] = 1;
      do_erase(id, 3);
    } else {
      for (long long i = 0; i < 5; ++i) seg[id].d[i][i] = 0;
    }
    return;
  }
  long long mid = (r + l) >> 1, lc = id << 1, rc = lc | 1;
  build(lc, l, mid);
  build(rc, mid, r);
  seg[id] = seg[lc] + seg[rc];
}
node get(long long x, long long y, long long id = 1, long long l = 0,
         long long r = n) {
  if (r <= x || l >= y) {
    node hichi;
    for (long long i = 0; i < 5; ++i) hichi.d[i][i] = 0;
    return hichi;
  }
  if (l >= x && r <= y) return seg[id];
  long long mid = (r + l) >> 1, lc = id << 1, rc = lc | 1;
  return (get(x, y, lc, l, mid) + get(x, y, rc, mid, r));
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> q >> s;
  build();
  while (q--) {
    long long l, r;
    cin >> l >> r;
    node res = get(--l, r);
    if (res.d[0][4] > n) res.d[0][4] = -1;
    cout << res.d[0][4] << endl;
  }
}

```