#include <bits/stdc++.h>
using namespace std;
static inline int count_bit(int x) {
  int ret = 0;
  for (; x; x -= x & -x) ret++;
  return ret;
}
static inline void _mod(long long &x) {
  if (x <= -1000000007 || x >= 1000000007) x %= 1000000007;
  if (x < 0) x += 1000000007;
}
static inline long long madd(long long x, long long y) {
  _mod(x);
  _mod(y);
  x += y;
  _mod(x);
  return x;
}
static inline long long msub(long long x, long long y) {
  _mod(x);
  _mod(y);
  x -= y;
  _mod(x);
  return x;
}
static inline long long mmul(long long x, long long y) {
  _mod(x);
  _mod(y);
  x *= y;
  _mod(x);
  return x;
}
static inline long long mpow(long long x, long long y) {
  long long ret = 1;
  for (; y; y >>= 1) {
    if (y & 1) ret = mmul(ret, x);
    x = mmul(x, x);
  }
  return ret;
}
long long n, m;
long long k;
long long l;
long long a[2000000];
long long dp[2][2000000], ps[2000000];
unordered_map<int, long long> f, f2;
int main() {
  long long ans = 0;
  scanf("%lld%lld%lld", &n, &l, &k);
  long long q = l / n, r = l % n;
  for (int i = (0); i < (n); i++) scanf("%lld", a + i);
  for (int i = (0); i < (r); i++) f2[a[i]]++;
  sort(a, a + n);
  for (int i = (0); i < (n); i++) f[a[i]]++;
  n = unique(a, a + n) - a;
  for (int i = (0); i < (n); i++) dp[1][i] = f[a[i]];
  for (int i = (0); i < (n); i++) ans = madd(ans, mmul(f[a[i]], q));
  for (int i = n - 1; i >= 0; i--) ps[i] = madd(f[a[i]], ps[i + 1]);
  for (int j = (2); j <= (min(k, q)); j++)
    for (int i = n - 1; i >= 0; i--) {
      dp[j & 1][i] = mmul(f[a[i]], ps[i]);
      ans = madd(ans, mmul(dp[j & 1][i], q - j + 1));
      ps[i] = madd(dp[j & 1][i], ps[i + 1]);
    }
  if (r) {
    memset(dp, 0, sizeof(dp));
    memset(ps, 0, sizeof(ps));
    for (int i = (0); i < (n); i++) dp[1][i] = f2[a[i]];
    for (int i = (0); i < (n); i++) ans = madd(ans, f2[a[i]]);
    for (int i = n - 1; i >= 0; i--) ps[i] = madd(f2[a[i]], ps[i + 1]);
    for (int j = (2); j <= (min(k, q + 1)); j++)
      for (int i = n - 1; i >= 0; i--) {
        dp[j & 1][i] = mmul(f[a[i]], ps[i]);
        ans = madd(ans, dp[j & 1][i]);
        ps[i] = madd(dp[j & 1][i], ps[i + 1]);
      }
  }
  printf("%lld\n", 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;
static inline int count_bit(int x) {
  int ret = 0;
  for (; x; x -= x & -x) ret++;
  return ret;
}
static inline void _mod(long long &x) {
  if (x <= -1000000007 || x >= 1000000007) x %= 1000000007;
  if (x < 0) x += 1000000007;
}
static inline long long madd(long long x, long long y) {
  _mod(x);
  _mod(y);
  x += y;
  _mod(x);
  return x;
}
static inline long long msub(long long x, long long y) {
  _mod(x);
  _mod(y);
  x -= y;
  _mod(x);
  return x;
}
static inline long long mmul(long long x, long long y) {
  _mod(x);
  _mod(y);
  x *= y;
  _mod(x);
  return x;
}
static inline long long mpow(long long x, long long y) {
  long long ret = 1;
  for (; y; y >>= 1) {
    if (y & 1) ret = mmul(ret, x);
    x = mmul(x, x);
  }
  return ret;
}
long long n, m;
long long k;
long long l;
long long a[2000000];
long long dp[2][2000000], ps[2000000];
unordered_map<int, long long> f, f2;
int main() {
  long long ans = 0;
  scanf("%lld%lld%lld", &n, &l, &k);
  long long q = l / n, r = l % n;
  for (int i = (0); i < (n); i++) scanf("%lld", a + i);
  for (int i = (0); i < (r); i++) f2[a[i]]++;
  sort(a, a + n);
  for (int i = (0); i < (n); i++) f[a[i]]++;
  n = unique(a, a + n) - a;
  for (int i = (0); i < (n); i++) dp[1][i] = f[a[i]];
  for (int i = (0); i < (n); i++) ans = madd(ans, mmul(f[a[i]], q));
  for (int i = n - 1; i >= 0; i--) ps[i] = madd(f[a[i]], ps[i + 1]);
  for (int j = (2); j <= (min(k, q)); j++)
    for (int i = n - 1; i >= 0; i--) {
      dp[j & 1][i] = mmul(f[a[i]], ps[i]);
      ans = madd(ans, mmul(dp[j & 1][i], q - j + 1));
      ps[i] = madd(dp[j & 1][i], ps[i + 1]);
    }
  if (r) {
    memset(dp, 0, sizeof(dp));
    memset(ps, 0, sizeof(ps));
    for (int i = (0); i < (n); i++) dp[1][i] = f2[a[i]];
    for (int i = (0); i < (n); i++) ans = madd(ans, f2[a[i]]);
    for (int i = n - 1; i >= 0; i--) ps[i] = madd(f2[a[i]], ps[i + 1]);
    for (int j = (2); j <= (min(k, q + 1)); j++)
      for (int i = n - 1; i >= 0; i--) {
        dp[j & 1][i] = mmul(f[a[i]], ps[i]);
        ans = madd(ans, dp[j & 1][i]);
        ps[i] = madd(dp[j & 1][i], ps[i + 1]);
      }
  }
  printf("%lld\n", ans);
  return 0;
}

```