#include <bits/stdc++.h>
using namespace std;
const int M = 1e6 + 10, mod = 1e9 + 7;
long long l;
int n, k;
long long a[M], b[M];
int mini[M];
int pre[M][2];
long long qkadd(long long a, long long b) {
  long long ret = 0;
  while (b) {
    if (b & 1) ret = (ret + a) % mod;
    b >>= 1;
    a = (a + a) % mod;
  }
  return ret;
}
long long solve() {
  vector<vector<int> > dp(k + 1, vector<int>(n + 1, 0));
  for (int i = 0; i < n; i++) {
    dp[1][i] = 1;
    pre[mini[i]][0] = mini[i];
  }
  for (int i = 2; i <= k; i++) {
    for (int j = 0; j <= n; j++) pre[j][1] = 0;
    for (int j = 0; j < n; j++) {
      dp[i][j] = pre[mini[j]][0];
      pre[mini[j]][1] = (pre[mini[j]][1] + dp[i][j]) % mod;
    }
    for (int j = 1; j <= n; j++) {
      pre[j][1] = (pre[j - 1][1] + pre[j][1]) % mod;
      pre[j][0] = pre[j][1];
    }
  }
  int tmp = k;
  k = min(1ll * k, l / n);
  long long sum = 0;
  for (int i = 1; i <= k; i++) {
    if (l / n - i + 1 == 0) break;
    for (int j = 0; j < n; j++) {
      sum = (sum + qkadd(dp[i][j], l / n - i + 1)) % mod;
    }
  }
  int r = l % n;
  if (r == 0) return sum;
  k = min(1ll * tmp, l / n + (l % n == 0 ? 0 : 1));
  for (int i = 0; i < r; i++) {
    for (int j = 1; j <= k; j++) {
      sum = (sum + dp[j][i]) % mod;
    }
  }
  return sum;
}
int main() {
  scanf("%d%I64d%d", &n, &l, &k);
  for (int i = 0; i < n; i++) {
    scanf("%I64d", a + i);
    b[i] = a[i];
  }
  sort(b, b + n);
  for (int i = 0; i < n; i++) {
    mini[i] = upper_bound(b, b + n, a[i]) - b;
  }
  printf("%I64d\n", solve());
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

While Duff was resting in the beach, she accidentally found a strange array b0, b1, ..., bl - 1 consisting of l positive integers. This array was strange because it was extremely long, but there was another (maybe shorter) array, a0, ..., an - 1 that b can be build from a with formula: bi = ai mod n where a mod b denoted the remainder of dividing a by b.

<image>

Duff is so curious, she wants to know the number of subsequences of b like bi1, bi2, ..., bix (0 ≤ i1 < i2 < ... < ix < l), such that: 

  * 1 ≤ x ≤ k
  * For each 1 ≤ j ≤ x - 1, <image>
  * For each 1 ≤ j ≤ x - 1, bij ≤ bij + 1. i.e this subsequence is non-decreasing. 



Since this number can be very large, she want to know it modulo 109 + 7.

Duff is not a programmer, and Malek is unavailable at the moment. So she asked for your help. Please tell her this number.

Input

The first line of input contains three integers, n, l and k (1 ≤ n, k, n × k ≤ 106 and 1 ≤ l ≤ 1018).

The second line contains n space separated integers, a0, a1, ..., an - 1 (1 ≤ ai ≤ 109 for each 0 ≤ i ≤ n - 1). 

Output

Print the answer modulo 1 000 000 007 in one line.

Examples

Input

3 5 3
5 9 1


Output

10


Input

5 10 3
1 2 3 4 5


Output

25

Note

In the first sample case, <image>. So all such sequences are: <image>, <image>, <image>, <image>, <image>, <image>, <image>, <image>, <image> and <image>.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int M = 1e6 + 10, mod = 1e9 + 7;
long long l;
int n, k;
long long a[M], b[M];
int mini[M];
int pre[M][2];
long long qkadd(long long a, long long b) {
  long long ret = 0;
  while (b) {
    if (b & 1) ret = (ret + a) % mod;
    b >>= 1;
    a = (a + a) % mod;
  }
  return ret;
}
long long solve() {
  vector<vector<int> > dp(k + 1, vector<int>(n + 1, 0));
  for (int i = 0; i < n; i++) {
    dp[1][i] = 1;
    pre[mini[i]][0] = mini[i];
  }
  for (int i = 2; i <= k; i++) {
    for (int j = 0; j <= n; j++) pre[j][1] = 0;
    for (int j = 0; j < n; j++) {
      dp[i][j] = pre[mini[j]][0];
      pre[mini[j]][1] = (pre[mini[j]][1] + dp[i][j]) % mod;
    }
    for (int j = 1; j <= n; j++) {
      pre[j][1] = (pre[j - 1][1] + pre[j][1]) % mod;
      pre[j][0] = pre[j][1];
    }
  }
  int tmp = k;
  k = min(1ll * k, l / n);
  long long sum = 0;
  for (int i = 1; i <= k; i++) {
    if (l / n - i + 1 == 0) break;
    for (int j = 0; j < n; j++) {
      sum = (sum + qkadd(dp[i][j], l / n - i + 1)) % mod;
    }
  }
  int r = l % n;
  if (r == 0) return sum;
  k = min(1ll * tmp, l / n + (l % n == 0 ? 0 : 1));
  for (int i = 0; i < r; i++) {
    for (int j = 1; j <= k; j++) {
      sum = (sum + dp[j][i]) % mod;
    }
  }
  return sum;
}
int main() {
  scanf("%d%I64d%d", &n, &l, &k);
  for (int i = 0; i < n; i++) {
    scanf("%I64d", a + i);
    b[i] = a[i];
  }
  sort(b, b + n);
  for (int i = 0; i < n; i++) {
    mini[i] = upper_bound(b, b + n, a[i]) - b;
  }
  printf("%I64d\n", solve());
  return 0;
}

```