CF305D.Olya and Graph

Olya has got a directed non-weighted graph, consisting of $n$ vertexes and $m$ edges. We will consider that the graph vertexes are indexed from 1 to $n$ in some manner. Then for any graph edge that goes from vertex $v$ to vertex $u$ the following inequation holds: v<u .

Now Olya wonders, how many ways there are to add an arbitrary (possibly zero) number of edges to the graph so as the following conditions were met:

1. You can reach vertexes number $i+1,i+2,...,n$ from any vertex number $i$ (i<n) .
2. For any graph edge going from vertex $v$ to vertex $u$ the following inequation fulfills: v<u .
3. There is at most one edge between any two vertexes.
4. The shortest distance between the pair of vertexes $i,j$ (i<j) , for which $j-i<=k$ holds, equals $j-i$ edges.
5. The shortest distance between the pair of vertexes $i,j$ (i<j) , for which j-i>k holds, equals either $j-i$ or $j-i-k$ edges.

We will consider two ways distinct, if there is the pair of vertexes $i,j$ (i<j) , such that first resulting graph has an edge from $i$ to $j$ and the second one doesn't have it.

Help Olya. As the required number of ways can be rather large, print it modulo $1000000007$ $(10^{9}+7)$ .

The first line contains three space-separated integers $n,m,k$ $(2<=n<=10^{6},0<=m<=10^{5},1<=k<=10^{6})$ .

The next $m$ lines contain the description of the edges of the initial graph. The $i$ -th line contains a pair of space-separated integers $u_{i},v_{i}$ (1<=u_{i}<v_{i}<=n) — the numbers of vertexes that have a directed edge from $u_{i}$ to $v_{i}$ between them.

It is guaranteed that any pair of vertexes $u_{i},v_{i}$ has at most one edge between them. It also is guaranteed that the graph edges are given in the order of non-decreasing $u_{i}$ . If there are multiple edges going from vertex $u_{i}$ , then it is guaranteed that these edges are given in the order of increasing $v_{i}$ .

Print a single integer — the answer to the problem modulo $1000000007$ $(10^{9}+7)$ .

In the first sample there are two ways: the first way is not to add anything, the second way is to add a single edge from vertex $2$ to vertex $5$ .