Bob came to a cash & carry store, put n items into his trolley, and went to the checkout counter to pay. Each item is described by its price ci and time ti in seconds that a checkout assistant spends on this item. While the checkout assistant is occupied with some item, Bob can steal some other items from his trolley. To steal one item Bob needs exactly 1 second. What is the minimum amount of money that Bob will have to pay to the checkout assistant? Remember, please, that it is Bob, who determines the order of items for the checkout assistant.

Input

The first input line contains number n (1 ≤ n ≤ 2000). In each of the following n lines each item is described by a pair of numbers ti, ci (0 ≤ ti ≤ 2000, 1 ≤ ci ≤ 109). If ti is 0, Bob won't be able to steal anything, while the checkout assistant is occupied with item i.

Output

Output one number — answer to the problem: what is the minimum amount of money that Bob will have to pay.

Examples

Input

4
2 10
0 20
1 5
1 3


Output

8


Input

3
0 1
0 10
0 100


Output

111
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
long long int invmod(long long int x, long long int n, long long int mod) {
  if (n == 0) return 1 % mod;
  long long int half = invmod(x, n / 2, mod);
  half = (half * half) % mod;
  if (n % 2 == 1) half = (half * (x % mod)) % mod;
  return half;
}
long long int bin(vector<long long int> &a, long long int l, long long int r,
                  long long int x) {
  if (l <= r) {
    long long int m = l + (r - l) / 2;
    if (a[m] == x)
      return m;
    else if (a[m] > x)
      return bin(a, l, m - 1, x);
    else
      return bin(a, m + 1, r, x);
  } else
    return -1;
}
long long int power(long long int b, long long int e, long long int m) {
  if (e == 0) return 1;
  if (e % 2)
    return b * power(b * b % m, (e - 1) / 2, m) % m;
  else
    return power(b * b % m, e / 2, m);
}
long long int power(long long int b, long long int e) {
  if (e == 0) return 1;
  if (e % 2)
    return b * power(b * b, (e - 1) / 2);
  else
    return power(b * b, e / 2);
}
long long int ncr(long long int n, long long int p) {
  long long int r = min(p, n - p), rf = 1, ln = 1;
  for (long long int i = 1; i <= r; i++) rf = rf * i;
  for (long long int i = 0; i < r; i++) ln = ln * (n - i);
  return ln / rf;
}
bool sbs(pair<long long int, long long int> &a,
         pair<long long int, long long int> &b) {
  return (a.second < b.second);
}
bool sbds(pair<long long int, long long int> &a,
          pair<long long int, long long int> &b) {
  return (a.second > b.second);
}
long long int SUM(vector<long long int> &a, long long int n) {
  long long int sum = 0;
  for (long long int i = 0; i < n; i++) sum += a[i];
  return sum;
}
long long int chkprm(long long int n) {
  long long int x = 5, s = sqrt(n);
  if (n < 2) return 0;
  if (n < 4) return 1;
  if ((n & 1) == 0) return 0;
  if (n % 3 == 0) return 0;
  while (x <= s) {
    if (n % x == 0) return 0;
    x += 2;
    if (n % x == 0) return 0;
    x += 4;
  }
  return 1;
}
long long int dp[2002][4002];
long long int solve(vector<long long int> &t, vector<long long int> &c,
                    long long int n, long long int idx, long long int sum) {
  if (dp[idx][sum] != -1) return dp[idx][sum];
  if (idx == n) {
    if (sum < n)
      return dp[idx][sum] = 1000000007;
    else
      return dp[idx][sum] = 0;
  }
  if (sum < n)
    return dp[idx][sum] = min(c[idx] + solve(t, c, n, idx + 1, sum + t[idx]),
                              solve(t, c, n, idx + 1, sum));
  else
    return dp[idx][sum] = solve(t, c, n, idx + 1, sum);
}
void myth() {
  long long int n;
  cin >> n;
  vector<long long int> t(n), c(n);
  for (long long int i = 0; i < n; i++) {
    cin >> t[i] >> c[i];
    t[i]++;
  }
  memset(dp, -1, sizeof(dp));
  cout << solve(t, c, n, 0, 0);
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int test_case = 1;
  while (test_case--) {
    myth();
    cout << "\n";
  }
  cerr << "\n"
       << "Time Elasped : " << 1.0 * clock() / CLOCKS_PER_SEC << "\n";
  return 0;
}
