Problem: You're given an array a_1, …, a_n of n non-negative integers.

Let's call it sharpened if and only if there exists an integer 1 ≤ k ≤ n such that a_1 < a_2 < … < a_k and a_k > a_{k+1} > … > a_n. In particular, any strictly increasing or strictly decreasing array is sharpened. For example:

  * The arrays [4], [0, 1], [12, 10, 8] and [3, 11, 15, 9, 7, 4] are sharpened; 
  * The arrays [2, 8, 2, 8, 6, 5], [0, 1, 1, 0] and [2, 5, 6, 9, 8, 8] are not sharpened. 



You can do the following operation as many times as you want: choose any strictly positive element of the array, and decrease it by one. Formally, you can choose any i (1 ≤ i ≤ n) such that a_i>0 and assign a_i := a_i - 1.

Tell if it's possible to make the given array sharpened using some number (possibly zero) of these operations.

Input

The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 15\ 000) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer n (1 ≤ n ≤ 3 ⋅ 10^5).

The second line of each test case contains a sequence of n non-negative integers a_1, …, a_n (0 ≤ a_i ≤ 10^9).

It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.

Output

For each test case, output a single line containing "Yes" (without quotes) if it's possible to make the given array sharpened using the described operations, or "No" (without quotes) otherwise.

Example

Input


10
1
248618
3
12 10 8
6
100 11 15 9 7 8
4
0 1 1 0
2
0 0
2
0 1
2
1 0
2
1 1
3
0 1 0
3
1 0 1


Output


Yes
Yes
Yes
No
No
Yes
Yes
Yes
Yes
No

Note

In the first and the second test case of the first test, the given array is already sharpened.

In the third test case of the first test, we can transform the array into [3, 11, 15, 9, 7, 4] (decrease the first element 97 times and decrease the last element 4 times). It is sharpened because 3 < 11 < 15 and 15 > 9 > 7 > 4.

In the fourth test case of the first test, it's impossible to make the given array sharpened.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T>
void read(T &x) {
  cin >> x;
}
template <typename T, typename T0>
void read(T &x, T0 &y) {
  cin >> x >> y;
}
template <typename T, typename T0, typename T1>
void read(T &x, T0 &y, T1 &z) {
  cin >> x >> y >> z;
}
template <typename T, typename T0, typename T1, typename T2>
void read(T &x, T0 &y, T1 &z, T2 &w) {
  cin >> x >> y >> z >> w;
}
template <typename T, typename T0>
void read(pair<T, T0> &p) {
  cin >> p.fst >> p.scd;
}
template <typename T>
void read(vector<T> &oneD) {
  for (long long int i = 0; i < (int)(oneD).size(); i++) {
    read(oneD[i]);
  }
}
template <typename T>
void read(T oneD[], long long int n) {
  for (long long int i = 0; i < n; i++) {
    read(oneD[i]);
  }
}
template <typename T>
void write(T &x) {
  cout << x << " ";
}
template <typename T, typename T0>
void write(T &x, T0 &y) {
  cout << x << " " << y << "\n";
}
template <typename T, typename T0, typename T1>
void write(T &x, T0 &y, T1 &z) {
  cout << x << " " << y << " " << z << "\n";
}
template <typename T, typename T0, typename T1, typename T2>
void write(T &x, T0 &y, T1 &z, T2 &w) {
  cout << x << " " << y << " " << z << " " << w << "\n";
}
template <typename T, typename T0>
void write(pair<T, T0> &p) {
  write(p.fst);
  write(p.scd);
  cout << endl;
}
template <typename T>
void write(vector<T> &oneD) {
  for (long long int i = 0; i < (int)(oneD).size(); i++) {
    write(oneD[i]);
  }
  cout << endl;
}
template <typename T>
void write(T oneD[], int n) {
  for (long long int i = 0; i < n; i++) {
    write(oneD[i]);
  }
  cout << endl;
}
template <typename T, typename T0>
void write(map<T, T0> &mpp) {
  for (auto it : mpp) {
    write(it.fst);
    cout << ": ";
    write(it.scd);
    cout << "\n";
  }
  cout << endl;
}
template <typename T>
inline bool umax(T &x, T y) {
  return (y > x) ? x = y, true : false;
}
template <typename T>
inline bool umin(T &x, T y) {
  return (y < x) ? x = y, true : false;
}
template <class T>
void offset(long long int o, T &x) {
  x += o;
}
template <class T>
void offset(long long int o, vector<T> &x) {
  for (auto &a : x) offset(o, a);
}
template <class T, size_t S>
void offset(long long int o, array<T, S> &x) {
  for (auto &a : x) offset(o, a);
}
inline bool isSquare(long long int x) {
  long long int s = sqrt(x);
  return (s * s == x);
}
inline bool isPowerOfTwo(long long int x) {
  return ((1LL << (long long int)log2(x)) == x);
}
long long int extended_GCD(long long int a, long long int b, long long int &x,
                           long long int &y) {
  if (a == 0) {
    x = 0;
    y = 1;
    return b;
  }
  long long int x1, y1;
  long long int gcd = extended_GCD(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return gcd;
}
vector<long long int> prime;
void sieve(long long int N) {
  bool isPrime[N + !1];
  for (long long int i = 0; i <= N; ++i) {
    isPrime[i] = true;
  }
  isPrime[0] = false;
  isPrime[1] = false;
  for (long long int i = 2; i * i <= N; ++i) {
    if (isPrime[i] == true) {
      for (long long int j = i * i; j <= N; j += i) isPrime[j] = false;
    }
  }
  for (long long int i = 2; i <= N; i++) {
    if (isPrime[i] == true) {
      prime.push_back(i);
    }
  }
}
long long int minPrime[(int)(1 * 1e6 + 10)];
void factorSieve(long long int n) {
  memset(minPrime, 0, sizeof(minPrime));
  for (long long int i = 2; i * i <= n; ++i) {
    if (minPrime[i] == 0) {
      for (long long int j = i * i; j <= n; j += i) {
        if (minPrime[j] == 0) {
          minPrime[j] = i;
        }
      }
    }
  }
  for (long long int i = 2; i <= n; ++i) {
    if (minPrime[i] == 0) {
      minPrime[i] = i;
    }
  }
}
long long int factorize(long long int n) {
  map<long long int, long long int> m;
  m.clear();
  while (n != 1) {
    m[minPrime[n]]++;
    n /= minPrime[n];
  }
  for (auto i : m) {
    if (i.second > 1) {
      return 0;
    }
  }
  return 1;
}
string to_string(char c) { return string(1, c); }
const long long int N = 5e4;
void solve() {
  long long int n;
  read(n);
  vector<long long int> arr;
  for (long long int i = (0); ((1) > 0 ? i < (n) : i > (n)); i += (1)) {
    long long int a;
    read(a);
    arr.push_back(a);
  }
  long long int pos = n, flag = 0;
  long long int inc[n], dec[n];
  for (long long int i = (0); ((1) > 0 ? i < (n) : i > (n)); i += (1)) {
    inc[i] = i;
  }
  for (long long int i = (n - 1); ((-1) > 0 ? i < (-1) : i > (-1)); i += (-1)) {
    dec[i] = (n - 1 - i);
  }
  for (long long int i = (0); ((1) > 0 ? i < (n) : i > (n)); i += (1)) {
    if (arr[i] < inc[i]) {
      pos = i - 1;
      break;
    }
  }
  for (long long int i = (pos); ((1) > 0 ? i < (n) : i > (n)); i += (1)) {
    if (arr[i] < dec[i]) flag = 1;
  }
  if (flag) {
    cout << "No";
  } else {
    cout << "Yes";
  }
  cout << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int t = 1;
  read(t);
  for (long long int i = (0); ((1) > 0 ? i < (t) : i > (t)); i += (1)) {
    solve();
  }
}