#include <bits/stdc++.h>
using namespace std;
template <typename T>
T gcd(T a, T b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long isPowerof2(long long x) { return (x && !(x & x - 1)); }
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ostream_iterator<long long> output(cout, " ");
  long long n;
  cin >> n;
  vector<long long> v[n];
  string second;
  cin >> second;
  for (long long(i) = (0); i < (n); i++) {
    v[i].emplace_back(second[i] - '0');
  }
  for (long long(i) = (0); i < (n); i++) {
    long long a, b;
    cin >> a >> b;
    long long ini = v[i][0];
    long long cur = ini;
    for (long long t = 1; t <= 10000; t++) {
      if (t < b) {
        v[i].emplace_back(ini);
      } else {
        long long tmp = t - b;
        if (tmp % a == 0) {
          cur = 1 ^ cur;
        }
        v[i].emplace_back(cur);
      }
    }
  }
  long long ans = -1;
  for (long long t = 0; t <= 10000; t++) {
    long long res = 0;
    for (long long i = 0; i < n; i++) {
      if (v[i][t]) {
        res++;
      }
    }
    ans = max(res, ans);
  }
  cout << ans;
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

It is a holiday season, and Koala is decorating his house with cool lights! He owns n lights, all of which flash periodically.

After taking a quick glance at them, Koala realizes that each of his lights can be described with two parameters a_i and b_i. Light with parameters a_i and b_i will toggle (on to off, or off to on) every a_i seconds starting from the b_i-th second. In other words, it will toggle at the moments b_i, b_i + a_i, b_i + 2 ⋅ a_i and so on.

You know for each light whether it's initially on or off and its corresponding parameters a_i and b_i. Koala is wondering what is the maximum number of lights that will ever be on at the same time. So you need to find that out.

<image> Here is a graphic for the first example.

Input

The first line contains a single integer n (1 ≤ n ≤ 100), the number of lights.

The next line contains a string s of n characters. The i-th character is "1", if the i-th lamp is initially on. Otherwise, i-th character is "0".

The i-th of the following n lines contains two integers a_i and b_i (1 ≤ a_i, b_i ≤ 5) — the parameters of the i-th light.

Output

Print a single integer — the maximum number of lights that will ever be on at the same time.

Examples

Input


3
101
3 3
3 2
3 1


Output


2


Input


4
1111
3 4
5 2
3 1
3 2


Output


4


Input


6
011100
5 3
5 5
2 4
3 5
4 2
1 5


Output


6

Note

For first example, the lamps' states are shown in the picture above. The largest number of simultaneously on lamps is 2 (e.g. at the moment 2).

In the second example, all lights are initially on. So the answer is 4.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <typename T>
T gcd(T a, T b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long isPowerof2(long long x) { return (x && !(x & x - 1)); }
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ostream_iterator<long long> output(cout, " ");
  long long n;
  cin >> n;
  vector<long long> v[n];
  string second;
  cin >> second;
  for (long long(i) = (0); i < (n); i++) {
    v[i].emplace_back(second[i] - '0');
  }
  for (long long(i) = (0); i < (n); i++) {
    long long a, b;
    cin >> a >> b;
    long long ini = v[i][0];
    long long cur = ini;
    for (long long t = 1; t <= 10000; t++) {
      if (t < b) {
        v[i].emplace_back(ini);
      } else {
        long long tmp = t - b;
        if (tmp % a == 0) {
          cur = 1 ^ cur;
        }
        v[i].emplace_back(cur);
      }
    }
  }
  long long ans = -1;
  for (long long t = 0; t <= 10000; t++) {
    long long res = 0;
    for (long long i = 0; i < n; i++) {
      if (v[i][t]) {
        res++;
      }
    }
    ans = max(res, ans);
  }
  cout << ans;
  return 0;
}

```