#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:512000000")
using namespace std;
void solve(bool);
void precalc();
clock_t start;
int testNumber = 1;
bool todo = true;
int main() {
  start = clock();
  int t = 1;
  cout.sync_with_stdio(0);
  cin.tie(0);
  precalc();
  cout.precision(10);
  cout << fixed;
  int testNum = 1;
  while (t--) {
    solve(true);
    ++testNumber;
  }
  return 0;
}
template <typename T>
T binpow(T q, T w, T mod) {
  if (!w) return 1 % mod;
  if (w & 1) return q * 1LL * binpow(q, w - 1, mod) % mod;
  return binpow(q * 1LL * q % mod, w / 2, mod);
}
template <typename T>
T gcd(T q, T w) {
  while (w) {
    q %= w;
    swap(q, w);
  }
  return q;
}
template <typename T>
T lcm(T q, T w) {
  return q / gcd(q, w) * w;
}
void precalc() {}
const int INF = (int)1e9;
struct Node {
  std::array<std::array<int, 5>, 5> dp;
  Node() {
    for (int i = 0; i < 5; ++i) {
      for (int j = 0; j < 5; ++j) {
        dp[i][j] = INF;
      }
    }
    for (int i = 0; i < 5; ++i) {
      dp[i][i] = 0;
    }
  }
  Node(int num) {
    for (int i = 0; i < 5; ++i) {
      for (int j = 0; j < 5; ++j) {
        dp[i][j] = INF;
      }
    }
    for (int i = 0; i < 5; ++i) {
      dp[i][i] = 0;
    }
    if (num == -1) {
      dp[3][3] = 1;
      dp[4][4] = 1;
    } else if (num < 4) {
      dp[num][num + 1] = 0;
      dp[num][num] = 1;
    }
  }
};
Node merge(const Node& l, const Node& r) {
  Node res;
  for (int i = 0; i < 5; ++i) {
    for (int j = i; j < 5; ++j) {
      res.dp[i][j] = INF;
      for (int x = i; x <= j; ++x) {
        res.dp[i][j] = min(res.dp[i][j], l.dp[i][x] + r.dp[x][j]);
      }
    }
  }
  return res;
}
int shift = 1;
Node tree[2 * (1 << 18) + 5];
Node rmq(int l, int r) {
  if (l >= r) {
    return Node();
  }
  if (l & 1) {
    return merge(tree[l], rmq(l + 1, r));
  }
  if (r & 1) {
    return merge(rmq(l, r - 1), tree[r - 1]);
  }
  return rmq(l / 2, r / 2);
}
Node get_res(int l, int r) { return rmq(l + shift, r + shift); }
void solve(bool read) {
  int n, Q;
  cin >> n >> Q;
  string s;
  cin >> s;
  while (shift < s.length()) {
    shift *= 2;
  }
  for (int i = 0; i < s.length(); ++i) {
    int val;
    switch (s[i]) {
      case '2':
        val = 0;
        break;
      case '0':
        val = 1;
        break;
      case '1':
        val = 2;
        break;
      case '7':
        val = 3;
        break;
      case '6':
        val = -1;
        break;
      default:
        val = 4;
    }
    tree[i + shift] = Node(val);
  }
  for (int i = shift - 1; i > 0; --i) {
    tree[i] = merge(tree[2 * i], tree[2 * i + 1]);
  }
  for (int i = 0; i < Q; ++i) {
    int l, r;
    cin >> l >> r;
    --l;
    Node res = get_res(l, r);
    if (res.dp[0][4] > 1e8) {
      cout << "-1\n";
      continue;
    }
    cout << res.dp[0][4] << "\n";
  }
}

### Prompt

Your challenge is to write a Cpp solution to the following problem:

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>
#pragma comment(linker, "/STACK:512000000")
using namespace std;
void solve(bool);
void precalc();
clock_t start;
int testNumber = 1;
bool todo = true;
int main() {
  start = clock();
  int t = 1;
  cout.sync_with_stdio(0);
  cin.tie(0);
  precalc();
  cout.precision(10);
  cout << fixed;
  int testNum = 1;
  while (t--) {
    solve(true);
    ++testNumber;
  }
  return 0;
}
template <typename T>
T binpow(T q, T w, T mod) {
  if (!w) return 1 % mod;
  if (w & 1) return q * 1LL * binpow(q, w - 1, mod) % mod;
  return binpow(q * 1LL * q % mod, w / 2, mod);
}
template <typename T>
T gcd(T q, T w) {
  while (w) {
    q %= w;
    swap(q, w);
  }
  return q;
}
template <typename T>
T lcm(T q, T w) {
  return q / gcd(q, w) * w;
}
void precalc() {}
const int INF = (int)1e9;
struct Node {
  std::array<std::array<int, 5>, 5> dp;
  Node() {
    for (int i = 0; i < 5; ++i) {
      for (int j = 0; j < 5; ++j) {
        dp[i][j] = INF;
      }
    }
    for (int i = 0; i < 5; ++i) {
      dp[i][i] = 0;
    }
  }
  Node(int num) {
    for (int i = 0; i < 5; ++i) {
      for (int j = 0; j < 5; ++j) {
        dp[i][j] = INF;
      }
    }
    for (int i = 0; i < 5; ++i) {
      dp[i][i] = 0;
    }
    if (num == -1) {
      dp[3][3] = 1;
      dp[4][4] = 1;
    } else if (num < 4) {
      dp[num][num + 1] = 0;
      dp[num][num] = 1;
    }
  }
};
Node merge(const Node& l, const Node& r) {
  Node res;
  for (int i = 0; i < 5; ++i) {
    for (int j = i; j < 5; ++j) {
      res.dp[i][j] = INF;
      for (int x = i; x <= j; ++x) {
        res.dp[i][j] = min(res.dp[i][j], l.dp[i][x] + r.dp[x][j]);
      }
    }
  }
  return res;
}
int shift = 1;
Node tree[2 * (1 << 18) + 5];
Node rmq(int l, int r) {
  if (l >= r) {
    return Node();
  }
  if (l & 1) {
    return merge(tree[l], rmq(l + 1, r));
  }
  if (r & 1) {
    return merge(rmq(l, r - 1), tree[r - 1]);
  }
  return rmq(l / 2, r / 2);
}
Node get_res(int l, int r) { return rmq(l + shift, r + shift); }
void solve(bool read) {
  int n, Q;
  cin >> n >> Q;
  string s;
  cin >> s;
  while (shift < s.length()) {
    shift *= 2;
  }
  for (int i = 0; i < s.length(); ++i) {
    int val;
    switch (s[i]) {
      case '2':
        val = 0;
        break;
      case '0':
        val = 1;
        break;
      case '1':
        val = 2;
        break;
      case '7':
        val = 3;
        break;
      case '6':
        val = -1;
        break;
      default:
        val = 4;
    }
    tree[i + shift] = Node(val);
  }
  for (int i = shift - 1; i > 0; --i) {
    tree[i] = merge(tree[2 * i], tree[2 * i + 1]);
  }
  for (int i = 0; i < Q; ++i) {
    int l, r;
    cin >> l >> r;
    --l;
    Node res = get_res(l, r);
    if (res.dp[0][4] > 1e8) {
      cout << "-1\n";
      continue;
    }
    cout << res.dp[0][4] << "\n";
  }
}

```