#include <bits/stdc++.h>
using namespace std;
const int maxn = 2000005;
struct node {
  long long x, idx;
};
bool cmp(node x, node y) {
  if (x.x == y.x) return x.idx < y.idx;
  return x.x < y.x;
}
long long dp[2][maxn], sum[2][maxn], b[maxn];
node a[maxn];
int main() {
  long long n, l, k;
  scanf("%I64d%I64d%I64d", &n, &l, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%I64d", &a[i].x);
    a[i].idx = i;
  }
  sort(a + 1, a + 1 + n, cmp);
  for (int i = n; i >= 1; i--) {
    if (i == n) {
      b[i] = n;
      continue;
    }
    if (a[i].x == a[i + 1].x)
      b[i] = b[i + 1];
    else
      b[i] = i;
  }
  for (int i = 1; i <= n; i++) sum[1][i] = 1;
  long long ans = 0, now = 0, nxt = 1;
  for (long long i = 1; i <= k; i++) {
    if (l / n - i + 1 < 0) break;
    for (long long j = 1; j <= n; j++) {
      dp[now][j] = sum[nxt][b[j]];
      sum[now][j] = (sum[now][j - 1] + dp[now][j]) % 1000000007;
      long long tmp = l / n - i + 1;
      if (a[j].idx <= l % n) tmp++;
      ans = (ans + tmp % 1000000007 * dp[now][j] % 1000000007) % 1000000007;
    }
    swap(nxt, now);
  }
  printf("%I64d", ans);
  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 maxn = 2000005;
struct node {
  long long x, idx;
};
bool cmp(node x, node y) {
  if (x.x == y.x) return x.idx < y.idx;
  return x.x < y.x;
}
long long dp[2][maxn], sum[2][maxn], b[maxn];
node a[maxn];
int main() {
  long long n, l, k;
  scanf("%I64d%I64d%I64d", &n, &l, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%I64d", &a[i].x);
    a[i].idx = i;
  }
  sort(a + 1, a + 1 + n, cmp);
  for (int i = n; i >= 1; i--) {
    if (i == n) {
      b[i] = n;
      continue;
    }
    if (a[i].x == a[i + 1].x)
      b[i] = b[i + 1];
    else
      b[i] = i;
  }
  for (int i = 1; i <= n; i++) sum[1][i] = 1;
  long long ans = 0, now = 0, nxt = 1;
  for (long long i = 1; i <= k; i++) {
    if (l / n - i + 1 < 0) break;
    for (long long j = 1; j <= n; j++) {
      dp[now][j] = sum[nxt][b[j]];
      sum[now][j] = (sum[now][j - 1] + dp[now][j]) % 1000000007;
      long long tmp = l / n - i + 1;
      if (a[j].idx <= l % n) tmp++;
      ans = (ans + tmp % 1000000007 * dp[now][j] % 1000000007) % 1000000007;
    }
    swap(nxt, now);
  }
  printf("%I64d", ans);
  return 0;
}

```