#include "header.hpp"
#include <algorithm>
#include <cmath>
#include <map>
#include <utility>

int fact(int n) {
  int64_t result = 1;
  for (int i = 1; i <= n; ++i) {
    result *= i;
  }
  return result;
}

int selection(int n, int m) {
  return (fact(m) / (fact(n) * fact(std::abs(m - n))));
}

class Solution {
public:
  static int threeSumMulti(std::vector<int> &arr, int target) {
    int MOD = 1000000007;
    long ans = 0;
    std::sort(arr.begin(), arr.end());

    for (int i = 0; i < arr.size(); ++i) {
      // We'll try to find the number of i < j < k
      // with A[j] + A[k] == T, where T = target - A[i].

      // The below is a "two sum with multiplicity".
      int T = target - arr[i];
      int j = i + 1, k = arr.size() - 1;

      while (j < k) {
        // These steps proceed as in a typical two-sum.
        if (arr[j] + arr[k] < T)
          j++;
        else if (arr[j] + arr[k] > T)
          k--;
        else if (arr[j] != arr[k]) { // We have A[j] + A[k] == T.
          // Let's count "left": the number of A[j] == A[j+1] == A[j+2] == ...
          // And similarly for "right".
          int left = 1, right = 1;
          while (j + 1 < k && arr[j] == arr[j + 1]) {
            left++;
            j++;
          }
          while (k - 1 > j && arr[k] == arr[k - 1]) {
            right++;
            k--;
          }

          ans += left * right;
          ans %= MOD;
          j++;
          k--;
        } else {
          // M = k - j + 1
          // We contributed M * (M-1) / 2 pairs.
          ans += (k - j + 1) * (k - j) / 2;
          ans %= MOD;
          break;
        }
      }
    }

    return (int)ans;
  }
};

int main(int argc, char **argv) {
  std::vector<int> arr = {1, 1, 2, 2, 3, 3, 4, 4, 5, 5};
  auto target = 8;
  std::cout << Solution::threeSumMulti(arr, target) << "\n";
  return 0;
}