Let's call an array a consisting of n positive (greater than 0) integers beautiful if the following condition is held for every i from 1 to n: either a_i = 1, or at least one of the numbers a_i - 1 and a_i - 2 exists in the array as well.

For example: 

  * the array [5, 3, 1] is beautiful: for a_1, the number a_1 - 2 = 3 exists in the array; for a_2, the number a_2 - 2 = 1 exists in the array; for a_3, the condition a_3 = 1 holds; 
  * the array [1, 2, 2, 2, 2] is beautiful: for a_1, the condition a_1 = 1 holds; for every other number a_i, the number a_i - 1 = 1 exists in the array; 
  * the array [1, 4] is not beautiful: for a_2, neither a_2 - 2 = 2 nor a_2 - 1 = 3 exists in the array, and a_2 ≠ 1; 
  * the array [2] is not beautiful: for a_1, neither a_1 - 1 = 1 nor a_1 - 2 = 0 exists in the array, and a_1 ≠ 1; 
  * the array [2, 1, 3] is beautiful: for a_1, the number a_1 - 1 = 1 exists in the array; for a_2, the condition a_2 = 1 holds; for a_3, the number a_3 - 2 = 1 exists in the array. 



You are given a positive integer s. Find the minimum possible size of a beautiful array with the sum of elements equal to s.

Input

The first line contains one integer t (1 ≤ t ≤ 5000) — the number of test cases.

Then t lines follow, the i-th line contains one integer s (1 ≤ s ≤ 5000) for the i-th test case.

Output

Print t integers, the i-th integer should be the answer for the i-th testcase: the minimum possible size of a beautiful array with the sum of elements equal to s.

Example

Input


4
1
8
7
42


Output


1
3
3
7

Note

Consider the example test:

  1. in the first test case, the array [1] meets all conditions; 
  2. in the second test case, the array [3, 4, 1] meets all conditions; 
  3. in the third test case, the array [1, 2, 4] meets all conditions; 
  4. in the fourth test case, the array [1, 4, 6, 8, 10, 2, 11] meets all conditions. 
#include <bits/stdc++.h>
using namespace std;

#define pb              push_back
#define ll              long long
#define ld              long double
#define N               cout << "\n"
#define fr(i,n)         for(ll i=0;i<n ;i++)
#define fr1(i,n)        for(ll i=1;i<=n;i++)
#define fr2(n,m)        for(ll i=n;i<=m;i++)
#define co(x)           cout << x
#define cc(x)           cout << x << " "
#define cl(x)           cout << x << "\n"
#define coo(x,z)        cout << x << " " << z <<"\n";
#define cooo(x,z,k)     cout << x << " " << z << " " << k << "\n"
#define ms(x,z)         memset(x, z, sizeof(x))
#define reversed(s)     reverse(s.begin(), s.end())
#define asort(s)        sort(s.begin(), s.end())
#define dsort(s)        sort(s.begin(), s.end(),greater<ll>())
#define pi              pair <ll, ll>
#define ps              pair <string, ll>
#define x               first
#define z               second
#define bitcnt(n)       __builtin_popcntll(n)
#define all(s)          s.begin(), s.end()
void FAST() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL); cout.tie(NULL);
    #ifndef ONLINE_JUDGE
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
    #endif
}

const ll SZ = 500007;
const ll mod = 1000000007;                                                                                                                                                                                        ll Set(ll n,ll pos){ return n=n | (1LL<<pos); } ll totet(ll n,ll pos){ return n= n & ~(1LL<<pos); }bool check(ll n,ll pos){ return (bool)(n & (1LL<<pos)); }ll bigmod(ll x, ll y) { if(y == 0) return 1; ll tot = bigmod(x, y/2); tot = (tot *tot) % mod; if(y%2 == 1) tot = (tot * x) % mod; return tot;} ll par[SZ]; ll Find(ll node){ if(par[node]==node) return node; return par[node] = Find(par[node]);} void Union(ll a, ll b) { par[Find(b)] = Find(a); }

ll i,j,k,n,m,x,z,ans,q,cnt,l,d,r,t,cs,sz,a[SZ],b[SZ];
int main()
{
    FAST();
    cin >> t;
    while(t--){
        cin >> n;
        x = 1, k = 1, z = 1;
        while(k < n){
            x += 2;
            k += x;
            z++;
        }
        cl(z);
    }
    return 0;
}