There are red and blue balls in a basket. If one red and one blue ball are removed from this basket, then the number of red balls in the remaining balls is twice the number of blue balls. After this, 3 red and 3 blue balls are removed from the remaining balls. Now the last number of remaining red balls is found to be three times the last number of remaining blue balls. The number of blue balls in the start in the basket is

• A. 7
• B. 13
• C. 14
• D. 26