Shohag has an integer sequence a_1, a_2, …, a_n. He can perform the following operation any number of times (possibly, zero):

  * Select any positive integer k (it can be different in different operations). 
  * Choose any position in the sequence (possibly the beginning or end of the sequence, or in between any two elements) and insert k into the sequence at this position. 
  * This way, the sequence a changes, and the next operation is performed on this changed sequence. 



For example, if a=[3,3,4] and he selects k = 2, then after the operation he can obtain one of the sequences [\underline{2},3,3,4], [3,\underline{2},3,4], [3,3,\underline{2},4], or [3,3,4,\underline{2}].

Shohag wants this sequence to satisfy the following condition: for each 1 ≤ i ≤ |a|, a_i ≤ i. Here, |a| denotes the size of a.

Help him to find the minimum number of operations that he has to perform to achieve this goal. We can show that under the constraints of the problem it's always possible to achieve this goal in a finite number of operations.

Input

The first line contains a single integer t (1 ≤ t ≤ 200) — the number of test cases.

The first line of each test case contains a single integer n (1 ≤ n ≤ 100) — the initial length of the sequence.

The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — the elements of the sequence.

Output

For each test case, print a single integer — the minimum number of operations needed to perform to achieve the goal mentioned in the statement.

Example

Input


4
3
1 3 4
5
1 2 5 7 4
1
1
3
69 6969 696969


Output


1
3
0
696966

Note

In the first test case, we have to perform at least one operation, as a_2=3>2. We can perform the operation [1, 3, 4] → [1, \underline{2}, 3, 4] (the newly inserted element is underlined), now the condition is satisfied.

In the second test case, Shohag can perform the following operations:

[1, 2, 5, 7, 4] → [1, 2, \underline{3}, 5, 7, 4] → [1, 2, 3, \underline{4}, 5, 7, 4] → [1, 2, 3, 4, 5, \underline{3}, 7, 4].

In the third test case, the sequence already satisfies the condition.
1
39
481199252 336470888 634074578 642802746 740396295 773386884 579721198 396628655 503722503 971207868 202647942 2087506 268792718 46761498 443917727 16843338 125908043 691952768 717268783 787375312 150414369 693319712 519096230 45277106 856168102 762263554 674936674 407246545 274667941 279198849 527268921 155191459 421436316 536134606 286802932 140515506 646837311 233465964 451394766

971207858

/*بِسْمِ اللَّهِ الرَّحْمَنِ الرَّحِيم*/

//#pragma GCC optimize("O3,unroll-loops")
//#pragma GCC target("avx,avx2,fma")

#include <bits/stdc++.h>
using namespace std;
#define FASTIO ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
typedef long long ll;
using pii = pair<int, int>;
const double PI = acos(-1.0);
const ll mod = 1e9 + 7;
//const ll mod = 998244353;


inline void normal(ll &a) { a %= mod; (a < 0) && (a += mod); }
inline ll modMul(ll a, ll b) { a %= mod, b %= mod; normal(a), normal(b); return (a * b) % mod; }
inline ll modAdd(ll a, ll b) { a %= mod, b %= mod; normal(a), normal(b); return (a + b) % mod; }
inline ll modSub(ll a, ll b) { a %= mod, b %= mod; normal(a), normal(b); a -= b; normal(a); return a; }
inline ll modPow(ll b, ll p) { ll r = 1; while (p) { if (p & 1) r = modMul(r, b); b = modMul(b, b); p >>= 1; } return r; }
inline ll modInverse(ll a) { return modPow(a, mod - 2); }
inline ll modDiv(ll a, ll b) { return modMul(a, modInverse(b)); }



int dx[] = {1, -1, 0, 0};
int dy[] = {0, 0, 1, -1};
int dx8[] = {0, 0, 1, 1, 1, -1, -1, -1};
int dy8[] = {1, -1, -1, 0, 1, -1, 0, 1};
int kx8[] = {1, 1, 2, 2, -1, -1, -2, -2};
int ky8[] = {2, -2, 1, -1, 2, -2, 1, -1};
/* for Random Number generate
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
//*/
///**
template < typename F, typename S >ostream& operator << ( ostream& os, const pair< F, S > & p ) {return os << "(" << p.first << ", " << p.second << ")";}
template < typename T >ostream &operator << ( ostream & os, const vector< T > &v ) {os << "{"; for (auto it = v.begin(); it != v.end(); ++it) {if ( it != v.begin() ) os << ", "; os << *it;} return os << "}";}
template < typename T >ostream &operator << ( ostream & os, const set< T > &v ) {os << "["; for (auto it = v.begin(); it != v.end(); ++it) {if ( it != v.begin()) os << ", "; os << *it;} return os << "]";}
template < typename F, typename S >ostream &operator << ( ostream & os, const map< F, S > &v ) {os << "["; for (auto it = v.begin(); it != v.end(); ++it) {if ( it != v.begin() ) os << ", "; os << it -> first << " = " << it -> second ;} return os << "]";}
#define dbg(args...) do {cerr << #args << " : "; faltu(args); } while(0)
clock_t tStart = clock();
#define timeStamp dbg("Execution Time: ", (double)(clock() - tStart)/CLOCKS_PER_SEC)
void faltu () { cerr << endl; }
template <typename T>void faltu( T a[], int n ) {for (int i = 0; i < n; ++i) cerr << a[i] << ' '; cerr << endl;}
template <typename T, typename ... hello>
void faltu( T arg, const hello &... rest) { cerr << arg << ' '; faltu(rest...); }

// Program showing a policy-based data structure.
#include <ext/pb_ds/assoc_container.hpp> // Common file 
#include <ext/pb_ds/tree_policy.hpp>
#include <functional> // for less 
using namespace __gnu_pbds;

// GNU link : https://goo.gl/WVDL6g
typedef tree<int, null_type, less_equal<int>, rb_tree_tag,
        tree_order_statistics_node_update>
        new_data_set;
// find_by_order(k) – ফাংশনটি kth ordered element এর একটা পয়েন্টার রিটার্ন করে। অর্থাৎ তুমি চাইলেই kth ইন্ডেক্সে কি আছে, সেটা জেনে ফেলতে পারছো!
// order_of_key(x) – ফাংশনটি x এলিমেন্টটা কোন পজিশনে আছে সেটা বলে দেয়।
//*//**___________________________________________________**/
const int N = 1000006;


int main()
{
	FASTIO
	///*
#ifndef ONLINE_JUDGE
	freopen("in.txt", "r", stdin);
	freopen("out.txt", "w", stdout);
	freopen("error.txt", "w", stderr);
#endif
//*/
	int T;
	T = 1;
	//scanf("%d", &T);
	cin >> T;
	for (int cs = 1; cs <= T; cs++) {
		ll n;
		cin >> n;
		ll mx = 0;
		int pos = 0;
		for (int i = 0 ; i < n; i++) {
			ll x;
			cin >> x;
			if (mx < x) {
				mx = x;
				pos = i + 1;
			}
		}
		cout << max(mx - pos, 0ll) << "\n";
	}
	return 0;
}
/*
* use std::array instead of std::vector, if u can
* overflow?
* array bounds
*/