#include <stdio.h>

int main(){
    int t; 
    scanf("%d",&t);
    for(int a0 = 0; a0 < t; a0++){
        int n;
        scanf("%d",&n);
        n--;
        long long sum = 0;

        long long num3 = n / 3;
        long long num5 = n / 5;
        long long num15 = n / 15;

sum = 3 * num3 * (num3 + 1) / 2 + 5 * num5 * (num5 + 1) / 2 - 15 * num15 * (num15 + 1) / 2;

        printf("%lld\n", sum);
    }
    return 0;
}

private static long calculateSum(int n) {
    n--; // We want multiples below n, 
         // so decrement n by 1
    long sum = 0;

    if (n >= 3) {
        long mulThree = n / 3;
        sum += 3 * mulThree * (mulThree + 1) / 2;
    }

    if (n >= 5) {
        long mulFive = n / 5;
        sum += 5 * mulFive * (mulFive + 1) / 2;
    }
    
    // Remove duplicates
    if (n >= 15) {
        long mulFifteen = n / 15;
        sum -= 15 * mulFifteen * (mulFifteen + 1) / 2;
    }

    return sum;
}

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int t = in.nextInt();
        for(int a0 = 0; a0 < t; a0++){
            int n = in.nextInt();
            // Multiples under n, so decreasing n by 1
            n = n-1;
            // sum of all multiples of 3
            BigInteger p = BigInteger.valueOf(n/3);
            BigInteger s3 = BigInteger.valueOf(3).multiply(p).multiply(p.add(BigInteger.valueOf(1))).divide(BigInteger.valueOf(2));
            // sum of all multiples of 5
            p = BigInteger.valueOf(n/5);
            BigInteger s5 = BigInteger.valueOf(5).multiply(p).multiply(p.add(BigInteger.valueOf(1))).divide(BigInteger.valueOf(2));
            // sum of all multiples of 15
            p = BigInteger.valueOf(n/15);
            BigInteger s15 = BigInteger.valueOf(15).multiply(p).multiply(p.add(BigInteger.valueOf(1))).divide(BigInteger.valueOf(2));
            System.out.println(s3.add(s5).subtract(s15));
        }
        
        
    }
}

long long sumDivisableBy(long long num, long long target) {
    long long p = (--target) / num;
    return num * p * (p + 1) / 2;
}

int main() {
    int t;
    cin >> t;

    for (int a0 = 0; a0 < t; a0++) {
        int n;
        cin >> n;

        cout << sumDivisableBy(5, n) + sumDivisableBy(3, n) - sumDivisableBy(15, n) << endl;
    }
    return 0;
}