#include <bits/stdc++.h>
using namespace std;
const int maxn = 100 + 7;
const int ooi = 1000000 + 7;
const long long ool = 1e18 + 7;
const int apsz = 26;
const long long mod = 1e9 + 7;
int sta[maxn];
string s;
int n;
int a[maxn];
int b[maxn];
long long nowt;
int ok = 0;
int ans = 0;
int sta2[maxn];
int t[maxn];
int vis[maxn];
void init() {
  cin >> n;
  cin >> s;
  for (int i = (0); i < (n); ++i) sta2[i] = sta[i] = s[i] - '0';
  for (int i = (0); i < (n); ++i) cin >> a[i] >> b[i];
}
int over() {
  int tmp = 0;
  int o = 1;
  if (ok != n) o = 0;
  for (int i = (0); i < (n); ++i) {
    if (sta[i] == 1) tmp++;
    if (sta[i] != sta2[i]) o = 0;
  }
  ans = max(ans, tmp);
  return 0;
}
void solve() {
  while (!over() && nowt <= ooi) {
    nowt++;
    for (int i = (0); i < (n); ++i) {
      t[i]++;
      if (vis[i]) t[i] %= a[i];
      if (t[i] == b[i] && !vis[i]) {
        vis[i] = 1;
        t[i] = 0;
        ok++;
      }
      if (t[i] == 0 && vis[i]) sta[i] ^= 1;
    }
  }
  cout << ans << endl;
}
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  init();
  solve();
  return 0;
}

### Prompt

Create a solution in cpp for the following problem:

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;
const int maxn = 100 + 7;
const int ooi = 1000000 + 7;
const long long ool = 1e18 + 7;
const int apsz = 26;
const long long mod = 1e9 + 7;
int sta[maxn];
string s;
int n;
int a[maxn];
int b[maxn];
long long nowt;
int ok = 0;
int ans = 0;
int sta2[maxn];
int t[maxn];
int vis[maxn];
void init() {
  cin >> n;
  cin >> s;
  for (int i = (0); i < (n); ++i) sta2[i] = sta[i] = s[i] - '0';
  for (int i = (0); i < (n); ++i) cin >> a[i] >> b[i];
}
int over() {
  int tmp = 0;
  int o = 1;
  if (ok != n) o = 0;
  for (int i = (0); i < (n); ++i) {
    if (sta[i] == 1) tmp++;
    if (sta[i] != sta2[i]) o = 0;
  }
  ans = max(ans, tmp);
  return 0;
}
void solve() {
  while (!over() && nowt <= ooi) {
    nowt++;
    for (int i = (0); i < (n); ++i) {
      t[i]++;
      if (vis[i]) t[i] %= a[i];
      if (t[i] == b[i] && !vis[i]) {
        vis[i] = 1;
        t[i] = 0;
        ok++;
      }
      if (t[i] == 0 && vis[i]) sta[i] ^= 1;
    }
  }
  cout << ans << endl;
}
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  init();
  solve();
  return 0;
}

```