Codeforces Round 468 (Div. 1, based on Technocup 2018 Final Round)
E
Coins Exhibition
Codeforces
2900
原题
← D→
翻译中…

Arkady and Kirill visited an exhibition of rare coins. The coins were located in a row and enumerated from left to right from $1$ to $*k*$, each coin either was laid with its obverse (front) side up, or with its reverse (back) side up.

翻译中…

Arkady and Kirill made some photos of the coins, each photo contained a segment of neighboring coins. Akrady is interested in obverses, so on each photo made by him there is at least one coin with obverse side up. On the contrary, Kirill is interested in reverses, so on each photo made by him there is at least one coin with its reverse side up.

翻译中…

The photos are lost now, but Arkady and Kirill still remember the bounds of the segments of coins each photo contained. Given this information, compute the remainder of division by $10^{9} + 7$ of the number of ways to choose the upper side of each coin in such a way, that on each Arkady's photo there is at least one coin with obverse side up, and on each Kirill's photo there is at least one coin with reverse side up.

翻译中…

Input

翻译中…

The first line contains three integers $*k*$, $*n*$ and $*m*$ ($1 ≤ *k* ≤ 10^{9}$, $0 ≤ *n*, *m* ≤ 10^{5}$) — the total number of coins, the number of photos made by Arkady, and the number of photos made by Kirill, respectively.

翻译中…

The next $*n*$ lines contain the descriptions of Arkady's photos, one per line. Each of these lines contains two integers $*l*$ and $*r*$ ($1 ≤ *l* ≤ *r* ≤ *k*$), meaning that among coins from the $*l*$-th to the $*r*$-th there should be at least one with obverse side up.

翻译中…

The next $*m*$ lines contain the descriptions of Kirill's photos, one per line. Each of these lines contains two integers $*l*$ and $*r*$ ($1 ≤ *l* ≤ *r* ≤ *k*$), meaning that among coins from the $*l*$-th to the $*r*$-th there should be at least one with reverse side up.

翻译中…

Output

翻译中…

Print the only line — the number of ways to choose the side for each coin modulo $10^{9} + 7 = 1000000007$.

翻译中…

Note

翻译中…

In the first example the following ways are possible ('O' — obverse, 'R' — reverse side):

翻译中…
  • OROOR,
  • ORORO,
  • ORORR,
  • RROOR,
  • RRORO,
  • RRORR,
  • ORROR,
  • ORRRO.
翻译中…

In the second example the information is contradictory: the second coin should have obverse and reverse sides up at the same time, that is impossible. So, the answer is $0$.

翻译中…

样例

Input #1
5 2 2
1 3
3 5
2 2
4 5
Output #1
8
InputCopy OutputCopy
Input #2
5 3 2
1 3
2 2
3 5
2 2
4 5
Output #2
0
InputCopy OutputCopy
Input #3
60 5 7
1 3
50 60
1 60
30 45
20 40
4 5
6 37
5 18
50 55
22 27
25 31
44 45
Output #3
732658600
InputCopy OutputCopy
题目信息
题号
930E
时间限制
2 s
内存限制
256 MB
标签
数据结构
动态规划
数学

题面与样例版权归 Codeforces 及出题人所有;中文翻译由 AI 生成(CI 管线),如有出入请以原文为准。