计算K个反向对的数量时C++溢出

#include <iostream>
#include <vector>

using namespace std;

long long merge(vector<int>& nums, int left, int mid, int right) {
    vector<int> temp(right - left + 1);
    int i = left, j = mid + 1;
    long long count = 0;

    for (int k = 0; k < temp.size(); k++) {
        if (i > mid) {
            temp[k] = nums[j++];
        } else if (j > right) {
            temp[k] = nums[i++];
        } else if (nums[i] <= nums[j]) {
            temp[k] = nums[i++];
        } else {
            temp[k] = nums[j++];
            count += mid - i + 1;
        }
    }

    for (int k = 0; k < temp.size(); k++) {
        nums[left + k] = temp[k];
    }

    return count;
}

long long mergeSort(vector<int>& nums, int left, int right) {
    if (left >= right) {
        return 0;
    }

    int mid = left + (right - left) / 2;
    long long count = 0;

    count += mergeSort(nums, left, mid);
    count += mergeSort(nums, mid + 1, right);
    count += merge(nums, left, mid, right);

    return count;
}

long long reversePairs(vector<int>& nums) {
    return mergeSort(nums, 0, nums.size() - 1);
}

int main() {
    vector<int> nums = {1, 3, 2, 3, 1};
    long long count = reversePairs(nums);
    cout << "The number of reverse pairs is: " << count << endl;

    return 0;
}