if $ 100000 are invested in s&p500 in may 2011 , what will be the total value of the investment in may 2012?
Important information:
text_1: the line graph assumes the investment of $ 100 in our common stock , the standard & poor 2019s information technology index and the standard & poor 2019s 500 index on may 31 , 2011 and assumes reinvestment of all dividends .
table_1: the may 31 2011 of globalpayments is $ 100.00 ; the may 31 2011 of s&p500 is $ 100.00 ; the may 31 2011 of s&pinformationtechnology is $ 100.00 ;
table_2: the may 31 2012 of globalpayments is 81.90 ; the may 31 2012 of s&p500 is 99.59 ; the may 31 2012 of s&pinformationtechnology is 107.57 ;
Reasoning Steps:
Step: minus2-1(99.59, const_100) = 0.41
Step: divide2-2(#0, const_100) = 0.41%
Step: minus2-3(const_100, #1) = 99.59%
Step: multiply2-4(100000, #2) = 99590
Program:
subtract(99.59, const_100), divide(#0, const_100), subtract(const_100, #1), multiply(100000, #2)
Program (Nested):
multiply(100000, subtract(const_100, divide(subtract(99.59, const_100), const_100)))

Please answer the following financial question based on the context provided. Make sure to give the answer in raw number, not in percentage and without any units. If the question asks for percentage the value should be less than 1
Context:
stock performance graph the following graph compares our cumulative shareholder returns with the standard & poor 2019s information technology index and the standard & poor 2019s 500 index for the five years ended may 31 , 2016 . the line graph assumes the investment of $ 100 in our common stock , the standard & poor 2019s information technology index and the standard & poor 2019s 500 index on may 31 , 2011 and assumes reinvestment of all dividends . comparison of 5 year cumulative total return* among global payments inc. , the s&p 500 index and the s&p information technology index 5/11 5/145/135/12 global payments inc . s&p 500 s&p information technology 5/15 5/16 * $ 100 invested on may 31 , 2011 in stock or index , including reinvestment of dividends . fiscal year ending may 31 . copyright a9 2016 s&p , a division of the mcgraw-hill companies inc . all rights reserved . global payments information technology .
Table
            | globalpayments | s&p500   | s&pinformationtechnology
may 31 2011 | $ 100.00       | $ 100.00 | $ 100.00                
may 31 2012 | 81.90          | 99.59    | 107.57                  
may 31 2013 | 92.63          | 126.75   | 123.83                  
may 31 2014 | 132.59         | 152.67   | 153.42                  
may 31 2015 | 202.06         | 170.69   | 182.29                  
may 31 2016 | 300.97         | 173.62   | 187.97                  
issuer purchases of equity securities as announced on july 28 , 2015 , our board of directors authorized the additional repurchase of up to $ 300.0 million of our common stock . under these and other existing authorizations , we repurchased and retired 2.2 million shares of our common stock at a cost of $ 135.9 million including commissions , or an average price of $ 63.17 per share , during the year ended may 31 , 2016 . global payments inc . | 2016 form 10-k annual report 2013 29 .
Question: 
if $ 100000 are invested in s&p500 in may 2011 , what will be the total value of the investment in may 2012?
Important information:
text_1: the line graph assumes the investment of $ 100 in our common stock , the standard & poor 2019s information technology index and the standard & poor 2019s 500 index on may 31 , 2011 and assumes reinvestment of all dividends .
table_1: the may 31 2011 of globalpayments is $ 100.00 ; the may 31 2011 of s&p500 is $ 100.00 ; the may 31 2011 of s&pinformationtechnology is $ 100.00 ;
table_2: the may 31 2012 of globalpayments is 81.90 ; the may 31 2012 of s&p500 is 99.59 ; the may 31 2012 of s&pinformationtechnology is 107.57 ;
Reasoning Steps:
Step: minus2-1(99.59, const_100) = 0.41
Step: divide2-2(#0, const_100) = 0.41%
Step: minus2-3(const_100, #1) = 99.59%
Step: multiply2-4(100000, #2) = 99590
Program:
subtract(99.59, const_100), divide(#0, const_100), subtract(const_100, #1), multiply(100000, #2)
Program (Nested):
multiply(100000, subtract(const_100, divide(subtract(99.59, const_100), const_100)))
